Board Exam 2025
1 Mark
Q13. An optical device which always produces
images of m = + 1 is :
Magnification \(m = +1\) implies:
- Positive sign: Virtual and Erect.
- Magnitude 1: Same size as object.
This is the characteristic property of a Plane Mirror.
(Convex mirror always gives \(m < 1\)).
Correct Option: (A)
- Positive sign: Virtual and Erect.
- Magnitude 1: Same size as object.
This is the characteristic property of a Plane Mirror.
(Convex mirror always gives \(m < 1\)).
Correct Option: (A)
1 Mark
Q15. Concave mirror finds its application in
which of the following cases ?
(A) Solar Furnace: Concave mirrors converge sunlight to a
focus
to produce heat. Correct.
(B) Rear-view: Convex.
(C) Distant tall buildings: Convex.
(D) Divergent beam: Convex mirror / Concave lens.
Correct Option: (A)
(B) Rear-view: Convex.
(C) Distant tall buildings: Convex.
(D) Divergent beam: Convex mirror / Concave lens.
Correct Option: (A)
2 Marks
Q25. An object is placed 18 cm in front of a
concave mirror of focal length 12 cm. Use mirror formula to determine the position of the image
formed
in this case.
Given:
Object distance \(u = -18\) cm.
Focal length \(f = -12\) cm (Concave).
Mirror Formula: \(\frac{1}{v} + \frac{1}{u} = \frac{1}{f}\)
\(\frac{1}{v} = \frac{1}{f} - \frac{1}{u}\)
\(\frac{1}{v} = \frac{1}{-12} - \frac{1}{-18}\)
\(\frac{1}{v} = -\frac{1}{12} + \frac{1}{18}\)
LCM of 12 and 18 is 36.
\(\frac{1}{v} = \frac{-3 + 2}{36} = \frac{-1}{36}\)
\(v = -36\text{ cm}\)
Position: The image is formed 36 cm in front of the mirror (Real and
Object distance \(u = -18\) cm.
Focal length \(f = -12\) cm (Concave).
Mirror Formula: \(\frac{1}{v} + \frac{1}{u} = \frac{1}{f}\)
\(\frac{1}{v} = \frac{1}{f} - \frac{1}{u}\)
\(\frac{1}{v} = \frac{1}{-12} - \frac{1}{-18}\)
\(\frac{1}{v} = -\frac{1}{12} + \frac{1}{18}\)
LCM of 12 and 18 is 36.
\(\frac{1}{v} = \frac{-3 + 2}{36} = \frac{-1}{36}\)
\(v = -36\text{ cm}\)
Position: The image is formed 36 cm in front of the mirror (Real and
Q12. The magnification produced by a
rear-view
mirror fitted in vehicles :
Rear-view mirrors are Convex Mirrors.
Convex mirrors always form virtual, erect, and diminished (smaller) images.
Magnification (\(m = \frac{h'}{h}\)) is always less than one.
Correct Option: (A)
Convex mirrors always form virtual, erect, and diminished (smaller) images.
Magnification (\(m = \frac{h'}{h}\)) is always less than one.
Correct Option: (A)
1 Mark
Q13. A ray of light passes from a medium A to
another medium B. No refraction of light occurs if the ray of light hits the boundary of medium B at an
angle of :
No refraction occurs when light strikes the interface at a Normal
($$90^\circ$$ to the surface).
The angle of incidence is \(0^\circ\).
But the question asks "hits the boundary ... at an angle of". This usually refers to the angle made with the surface.
Angle with surface = \(90^\circ\).
Correct Option: (C)
The angle of incidence is \(0^\circ\).
But the question asks "hits the boundary ... at an angle of". This usually refers to the angle made with the surface.
Angle with surface = \(90^\circ\).
Correct Option: (C)
1 Mark
Q14. Which of the following ray diagrams is
correct for the ray of light incident on a lens as shown in figure ?
2025-31-2-QuestionNumber14
A ray passing through the Optical Centre of a convex lens goes
undeviated.
Fig showing straight path through O is correct.
Correct Option: (A) (Assuming Fig A shows undeviated path).
Fig showing straight path through O is correct.
Correct Option: (A) (Assuming Fig A shows undeviated path).
1 Mark
Q14. If the absolute refractive indices of two
media X and Y are \(\frac{6}{5}\) and \(\frac{4}{3}\) respectively, then the refractive index of Y with
respect to X will be :
Calculation:
\(n_X = \frac{6}{5}\). \(n_Y = \frac{4}{3}\).
Refractive index of Y w.r.t X (\(n_{YX}\)) = \(\frac{n_Y}{n_X}\).
\(n_{YX} = \frac{4/3}{6/5} = \frac{4}{3} \times \frac{5}{6} = \frac{20}{18} = \frac{10}{9}\).
Correct Option: (A)
\(n_X = \frac{6}{5}\). \(n_Y = \frac{4}{3}\).
Refractive index of Y w.r.t X (\(n_{YX}\)) = \(\frac{n_Y}{n_X}\).
\(n_{YX} = \frac{4/3}{6/5} = \frac{4}{3} \times \frac{5}{6} = \frac{20}{18} = \frac{10}{9}\).
Correct Option: (A)
1 Mark
Q15. An object is placed at a distance of 30 cm
from the pole of a concave mirror. If its real and inverted image is formed at 60 cm in front of the
mirror, the focal length of the mirror is :
Mirror Formula: \(\frac{1}{f} = \frac{1}{v} + \frac{1}{u}\)
\(u = -30 \text{ cm}\) (Object in front).
\(v = -60 \text{ cm}\) (Real image in front).
\(\frac{1}{f} = \frac{1}{-60} + \frac{1}{-30} = \frac{-1 - 2}{60} = \frac{-3}{60} = \frac{-1}{20}\).
\(f = -20 \text{ cm}\).
Correct Option: (B)
\(u = -30 \text{ cm}\) (Object in front).
\(v = -60 \text{ cm}\) (Real image in front).
\(\frac{1}{f} = \frac{1}{-60} + \frac{1}{-30} = \frac{-1 - 2}{60} = \frac{-3}{60} = \frac{-1}{20}\).
\(f = -20 \text{ cm}\).
Correct Option: (B)
2 Marks
Q25. (a) Out of the two lenses, one concave and
the other convex, state which one will diverge a parallel beam of light falling on it. Draw a ray
diagram to show the principal focus of the lens.
OR
(b) A ray of light after refraction from a convex lens emerges parallel to its principal axis.
(i) Draw a labelled ray diagram to show it.
(ii) In this case, the incident ray before refraction from the lens passes through a point on its principal axis. Name the point.
OR
(b) A ray of light after refraction from a convex lens emerges parallel to its principal axis.
(i) Draw a labelled ray diagram to show it.
(ii) In this case, the incident ray before refraction from the lens passes through a point on its principal axis. Name the point.
(a) Diverging Lens:
Concave Lens diverges a parallel beam.
Diagram: Parallel rays striking a concave lens and spreading out, appearing to diverge from the Principal Focus (F) on the same side.
(b) Convex Lens Case:
(i) Diagram: Rays coming from the Principal Focus (F) and emerging parallel to the axis.
(ii) Point Name: Principal Focus (F1).
Concave Lens diverges a parallel beam.
Diagram: Parallel rays striking a concave lens and spreading out, appearing to diverge from the Principal Focus (F) on the same side.
(b) Convex Lens Case:
(i) Diagram: Rays coming from the Principal Focus (F) and emerging parallel to the axis.
(ii) Point Name: Principal Focus (F1).
3 Marks
Q31. A convex mirror used for rear view on an
automobile has a focal length of 1.5 m. If a 3 m high bus is located at 6.0 m from the mirror, use
mirror formula to determine the position and size of the image of the bus as seen in the mirror.
Given:
\(f = +1.5 \text{ m}\) (Convex). \(u = -6.0 \text{ m}\) (Object). \(h_1 = 3 \text{ m}\).
Position (v):
\(\frac{1}{v} + \frac{1}{u} = \frac{1}{f}\)
\(\frac{1}{v} = \frac{1}{1.5} - \frac{1}{-6} = \frac{1}{1.5} + \frac{1}{6} = \frac{4}{6} + \frac{1}{6} = \frac{5}{6}\).
\(v = \frac{6}{5} = +1.2 \text{ m}\).
Image is 1.2 m behind the mirror.
Size (\(h_2\)):
\(m = \frac{h_2}{h_1} = \frac{-v}{u}\)
\(\frac{h_2}{3} = \frac{-1.2}{-6} = \frac{1.2}{6} = 0.2\).
\(h_2 = 3 \times 0.2 = 0.6 \text{ m}\).
Image is 0.6 m high (Diminished, Erect).
\(f = +1.5 \text{ m}\) (Convex). \(u = -6.0 \text{ m}\) (Object). \(h_1 = 3 \text{ m}\).
Position (v):
\(\frac{1}{v} + \frac{1}{u} = \frac{1}{f}\)
\(\frac{1}{v} = \frac{1}{1.5} - \frac{1}{-6} = \frac{1}{1.5} + \frac{1}{6} = \frac{4}{6} + \frac{1}{6} = \frac{5}{6}\).
\(v = \frac{6}{5} = +1.2 \text{ m}\).
Image is 1.2 m behind the mirror.
Size (\(h_2\)):
\(m = \frac{h_2}{h_1} = \frac{-v}{u}\)
\(\frac{h_2}{3} = \frac{-1.2}{-6} = \frac{1.2}{6} = 0.2\).
\(h_2 = 3 \times 0.2 = 0.6 \text{ m}\).
Image is 0.6 m high (Diminished, Erect).
2 Marks
Q21. An object is placed at a distance of 10 cm
in front of a concave mirror of focal length 15 cm. Use mirror formula to determine the position of the
image formed by this mirror.
Given:
\(u = -10\) cm, \(f = -15\) cm (Concave).
Mirror Formula: \(\frac{1}{v} + \frac{1}{u} = \frac{1}{f}\)
\(\frac{1}{v} = \frac{1}{-15} - \frac{1}{-10} = -\frac{1}{15} + \frac{1}{10}\)
LCM of 15 and 10 is 30.
\(\frac{1}{v} = \frac{-2 + 3}{30} = \frac{1}{30}\)
\(v = +30\) cm.
Position: Image is formed 30 cm behind the mirror (Virtual, Erect).
\(u = -10\) cm, \(f = -15\) cm (Concave).
Mirror Formula: \(\frac{1}{v} + \frac{1}{u} = \frac{1}{f}\)
\(\frac{1}{v} = \frac{1}{-15} - \frac{1}{-10} = -\frac{1}{15} + \frac{1}{10}\)
LCM of 15 and 10 is 30.
\(\frac{1}{v} = \frac{-2 + 3}{30} = \frac{1}{30}\)
\(v = +30\) cm.
Position: Image is formed 30 cm behind the mirror (Virtual, Erect).
3 Marks
Q29. Draw ray diagrams to show the nature,
position and relative size of the image formed by a convex mirror when the object is placed (i) at
infinity and (ii) between infinity and pole P of the mirror.
(i) At Infinity:
Position: At Focus (F) behind mirror. Size: Point-sized. Nature: Virtual & Erect.
(ii) Between Infinity and Pole:
Position: Between P and F behind mirror. Size: Diminished. Nature: Virtual & Erect.
Image: Convex Mirror Infinity
Position: At Focus (F) behind mirror. Size: Point-sized. Nature: Virtual & Erect.
(ii) Between Infinity and Pole:
Image: Convex Mirror Finite
Position: Between P and F behind mirror. Size: Diminished. Nature: Virtual & Erect.
5 Marks
Q34. (a) (i) The power of a lens ‘X’ is \(-2.5\)
D. Name the lens and determine its focal length in cm. For which eye defect of vision will an optician
prescribe this type of lens as a corrective lens ?
(ii) “The value of magnification ‘m’ for a lens is \(-2\).” Using new Cartesian Sign Convention and considering that an object is placed at a distance of 20 cm from the optical centre of this lens, state :
(I) the nature of the image formed;
(II) size of the image compared to the size of the object;
(III) position of the image, and
(IV) sign of the height of the image.
(iii) The numerical values of the focal lengths of two lenses A and B are 10 cm and 20 cm respectively. Which one of the two will show higher degree of convergence/divergence ? Give reason to justify your answer.
OR
(b) (i) Draw a ray diagram to show the refraction of a ray of light through a rectangular glass slab when it falls obliquely from air into glass.
(ii) State Snell’s law of refraction of light.
(iii) Differentiate between the virtual images formed by a convex lens and a concave lens on the basis of :
(I) object distance, and (II) magnification.
(ii) “The value of magnification ‘m’ for a lens is \(-2\).” Using new Cartesian Sign Convention and considering that an object is placed at a distance of 20 cm from the optical centre of this lens, state :
(I) the nature of the image formed;
(II) size of the image compared to the size of the object;
(III) position of the image, and
(IV) sign of the height of the image.
(iii) The numerical values of the focal lengths of two lenses A and B are 10 cm and 20 cm respectively. Which one of the two will show higher degree of convergence/divergence ? Give reason to justify your answer.
OR
(b) (i) Draw a ray diagram to show the refraction of a ray of light through a rectangular glass slab when it falls obliquely from air into glass.
(ii) State Snell’s law of refraction of light.
(iii) Differentiate between the virtual images formed by a convex lens and a concave lens on the basis of :
(I) object distance, and (II) magnification.
(a) (i) Lens X:
Power \(P = -2.5\) D (Negative power \(\implies\) Concave Lens).
Focal Length \(f = \frac{100}{P} = \frac{100}{-2.5} = -40\) cm.
Defect: Myopia (Near-sightedness).
(a) (ii) Magnification m = -2:
(I) Nature: Real and Inverted (Negative m).
(II) Size: Magnified (2 times object size).
(III) Position: \(m = \frac{v}{u} \implies -2 = \frac{v}{-20} \implies v = +40\) cm (Behind lens).
(IV) Sign of height: Negative (Inverted).
(a) (iii) Convergence/Divergence:
Power \(P \propto \frac{1}{f}\). Smaller focal length \(\implies\) Higher Power.
Lens A (10 cm) has higher power than Lens B (20 cm). So Lens A shows higher degree of convergence/divergence.
(b) (i) Glass Slab Ray Diagram:
(b) (ii) Snell's Law:
The ratio of sine of angle of incidence to the sine of angle of refraction is a constant for a given pair of media. \(\frac{\sin i}{\sin r} = n\).
(b) (iii) Differential:
1. Object Distance: Convex forms virtual image only when \(u < f\). Concave forms virtual image for all \(u\).
2. Magnification: Convex virtual image is magnified (\(m > 1\)). Concave virtual image is diminished (\(m < 1\)).
Power \(P = -2.5\) D (Negative power \(\implies\) Concave Lens).
Focal Length \(f = \frac{100}{P} = \frac{100}{-2.5} = -40\) cm.
Defect: Myopia (Near-sightedness).
(a) (ii) Magnification m = -2:
(I) Nature: Real and Inverted (Negative m).
(II) Size: Magnified (2 times object size).
(III) Position: \(m = \frac{v}{u} \implies -2 = \frac{v}{-20} \implies v = +40\) cm (Behind lens).
(IV) Sign of height: Negative (Inverted).
(a) (iii) Convergence/Divergence:
Power \(P \propto \frac{1}{f}\). Smaller focal length \(\implies\) Higher Power.
Lens A (10 cm) has higher power than Lens B (20 cm). So Lens A shows higher degree of convergence/divergence.
(b) (i) Glass Slab Ray Diagram:
Image: Glass Slab Refraction
(b) (ii) Snell's Law:
The ratio of sine of angle of incidence to the sine of angle of refraction is a constant for a given pair of media. \(\frac{\sin i}{\sin r} = n\).
(b) (iii) Differential:
1. Object Distance: Convex forms virtual image only when \(u < f\). Concave forms virtual image for all \(u\).
2. Magnification: Convex virtual image is magnified (\(m > 1\)). Concave virtual image is diminished (\(m < 1\)).
2 Marks
Q23. Draw a labeled ray diagram to show the
formation of image by a convex mirror when an object is placed at infinity. State the position and
size
of the image formed.
Ray Diagram (Object at Infinity, Convex Mirror):
Position: At the Focus (F) behind the mirror.
Size: Highly Diminished (Point-sized).
2025-31-2-Solution23
Position: At the Focus (F) behind the mirror.
Size: Highly Diminished (Point-sized).
5 Marks
Q38. (a) (i) "The linear magnification produced
by a spherical mirror is +3." Analyse this value and state the (I) type of mirror and (II) position of
the object with respect to the pole of the mirror.
(ii) Draw a labelled ray diagram to show the formation of image in this case.
OR
(b) (i) State the laws of refraction of light. Explain the term 'absolute refractive index of a medium' and write an expression to relate it with the speed of light in vacuum.
(ii) The absolute refractive indices of two media 'A' and 'B' are 2.0 and 1.5 respectively. If the speed of light in medium 'B' is \(2 \times 10^8\) m/s, calculate the speed of light in :
(I) vacuum
(II) medium 'A'
(ii) Draw a labelled ray diagram to show the formation of image in this case.
OR
(b) (i) State the laws of refraction of light. Explain the term 'absolute refractive index of a medium' and write an expression to relate it with the speed of light in vacuum.
(ii) The absolute refractive indices of two media 'A' and 'B' are 2.0 and 1.5 respectively. If the speed of light in medium 'B' is \(2 \times 10^8\) m/s, calculate the speed of light in :
(I) vacuum
(II) medium 'A'
(a) (i) Analysis of m = +3:
Positive sign indicates Virtual and Erect image.
Magnitude 3 (> 1) indicates Enlarged image.
(I) Type of Mirror: Concave Mirror (Only concave mirror forms enlarged virtual images).
(II) Object Position: Between Pole (P) and Focus (F).
(a) (ii) Ray Diagram:
r38.png"
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(b) (i) Laws of Refraction:
1. The incident ray, the refracted ray and the normal to the interface of two transparent media at the point of incidence, all lie in the same plane.
2. The ratio of sine of angle of incidence to the sine of angle of refraction is a constant (Snell's Law). \(\frac{\sin i}{\sin r} = \text{const}\).
Absolute Refractive Index: Ratio of speed of light in vacuum/air to the speed of light in the medium.
Expression: \(n_m = \frac{c}{v}\).
(b) (ii) Calculations:
Given: \(n_A = 2.0\), \(n_B = 1.5\), \(v_B = 2 \times 10^8\) m/s.
(I) Speed in Vacuum (c):
\(n_B = \frac{c}{v_B} \\Rightarrow c = n_B \times v_B\)
\(c = 1.5 \times 2 \times 10^8 = 3 \times 10^8\) m/s.
(II) Speed in Medium A (\(v_A\)):
\(n_A = \frac{c}{v_A} \\Rightarrow v_A = \frac{c}{n_A}\)
\(v_A = \frac{3 \times 10^8}{2.0} = 1.5 \times 10^8\) m/s.
Positive sign indicates Virtual and Erect image.
Magnitude 3 (> 1) indicates Enlarged image.
(I) Type of Mirror: Concave Mirror (Only concave mirror forms enlarged virtual images).
(II) Object Position: Between Pole (P) and Focus (F).
(a) (ii) Ray Diagram:
r38.png"
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object-fit: contain; height: auto; display: block; margin: 10px auto; border-radius: 8px;">(b) (i) Laws of Refraction:
1. The incident ray, the refracted ray and the normal to the interface of two transparent media at the point of incidence, all lie in the same plane.
2. The ratio of sine of angle of incidence to the sine of angle of refraction is a constant (Snell's Law). \(\frac{\sin i}{\sin r} = \text{const}\).
Absolute Refractive Index: Ratio of speed of light in vacuum/air to the speed of light in the medium.
Expression: \(n_m = \frac{c}{v}\).
(b) (ii) Calculations:
Given: \(n_A = 2.0\), \(n_B = 1.5\), \(v_B = 2 \times 10^8\) m/s.
(I) Speed in Vacuum (c):
\(n_B = \frac{c}{v_B} \\Rightarrow c = n_B \times v_B\)
\(c = 1.5 \times 2 \times 10^8 = 3 \times 10^8\) m/s.
(II) Speed in Medium A (\(v_A\)):
\(n_A = \frac{c}{v_A} \\Rightarrow v_A = \frac{c}{n_A}\)
\(v_A = \frac{3 \times 10^8}{2.0} = 1.5 \times 10^8\) m/s.
1 Mark
Q11. A candle flame is placed in front of the
reflecting surface of a convex mirror of focal length \(f\). If the distance of the flame from the pole
of the mirror is '\(f\)', its image is formed :
Analysis:
Mirror: Convex.
Object distance \(u = -f\) (in front).
Focal length \(f = +f\) (behind).
Mirror Formula: \(\frac{1}{v} + \frac{1}{u} = \frac{1}{f}\)
\(\frac{1}{v} = \frac{1}{f} - \frac{1}{u}\)
\(\frac{1}{v} = \frac{1}{f} - \frac{1}{-f} = \frac{1}{f} + \frac{1}{f} = \frac{2}{f}\)
\(v = \frac{f}{2}\).
Since \(v\) is positive, image is formed behind the mirror at distance \(\frac{f}{2}\).
Correct Option: (D)
Mirror: Convex.
Object distance \(u = -f\) (in front).
Focal length \(f = +f\) (behind).
Mirror Formula: \(\frac{1}{v} + \frac{1}{u} = \frac{1}{f}\)
\(\frac{1}{v} = \frac{1}{f} - \frac{1}{u}\)
\(\frac{1}{v} = \frac{1}{f} - \frac{1}{-f} = \frac{1}{f} + \frac{1}{f} = \frac{2}{f}\)
\(v = \frac{f}{2}\).
Since \(v\) is positive, image is formed behind the mirror at distance \(\frac{f}{2}\).
Correct Option: (D)
1 Mark
Q16. The image formed by a concave mirror is
observed to be real, inverted and larger than the object. Where is the object placed ?
Analysis:
Real, Inverted and Larger (Magnified) image is formed by a concave mirror when the object is placed between the Principal Focus (F) and Centre of Curvature (C).
(B) At C -> Same size.
(C) Beyond C -> Diminished.
(D) Between P and F -> Virtual, Erect, Magnified.
Correct Option: (A)
Real, Inverted and Larger (Magnified) image is formed by a concave mirror when the object is placed between the Principal Focus (F) and Centre of Curvature (C).
(B) At C -> Same size.
(C) Beyond C -> Diminished.
(D) Between P and F -> Virtual, Erect, Magnified.
Correct Option: (A)
2 Marks
Q21. Draw a ray diagram to show the path of the
reflected ray corresponding to an incident ray which is directed towards the principal focus of a convex
mirror. Mark the angle of incidence and angle of reflection on it.
Ray Diagram:
1. Incident ray is directed towards the Principal Focus (F) behind the mirror.
2. Rule: It gets reflected parallel to the principal axis.
3. Draw Normal (N) passing through Centre of Curvature (C).
4. Mark \( \angle i \) (between Incident Ray and Normal) and \( \angle r \) (between Reflected Ray and Normal).
1. Incident ray is directed towards the Principal Focus (F) behind the mirror.
2. Rule: It gets reflected parallel to the principal axis.
3. Draw Normal (N) passing through Centre of Curvature (C).
4. Mark \( \angle i \) (between Incident Ray and Normal) and \( \angle r \) (between Reflected Ray and Normal).
Diagram: Incident Ray towards F -> Parallel Reflected Ray
2 Marks
Q22. (a) Refractive index of diamond is 2.42.
What is the meaning of this statement in relation to speed of light ?
(b) Refractive index of water is 1.33. Calculate speed of light in water given speed of light in vacuum is \(3 \times 10^8\) m/s.
(b) Refractive index of water is 1.33. Calculate speed of light in water given speed of light in vacuum is \(3 \times 10^8\) m/s.
(a) Meaning:
It means that the speed of light in vacuum is 2.42 times the speed of light in diamond. (Or light travels 2.42 times slower in diamond than in vacuum).
Ratio \( \frac{c}{v} = 2.42 \).
(b) Calculation:
Given \( n_w = 1.33 \), \( c = 3 \times 10^8 \) m/s.
Formula: \( n = \frac{c}{v} \Rightarrow v = \frac{c}{n} \)
\( v = \frac{3 \times 10^8}{1.33} \approx 2.26 \times 10^8 \) m/s.
It means that the speed of light in vacuum is 2.42 times the speed of light in diamond. (Or light travels 2.42 times slower in diamond than in vacuum).
Ratio \( \frac{c}{v} = 2.42 \).
(b) Calculation:
Given \( n_w = 1.33 \), \( c = 3 \times 10^8 \) m/s.
Formula: \( n = \frac{c}{v} \Rightarrow v = \frac{c}{n} \)
\( v = \frac{3 \times 10^8}{1.33} \approx 2.26 \times 10^8 \) m/s.
5 Marks
Q35. (a) (i) "In refraction of light through a
rectangular glass slab, the emergent ray is always parallel to the direction of the incident ray." Why ?
Explain with the help of a ray diagram. What happens when a ray of light falls normally on one of the
faces of a rectangular glass prism ? Draw diagram.
(ii) An object is placed at a distance of 30 cm from the optical centre of a concave lens of focal length 20 cm. Use Lens formula to determine the position of the image formed in this case.
OR
(b) (i) A student wishes to study the image formation by a concave mirror using candle flame as object. State the type of the image formed by the mirror and mention the change in the image formed, if any, that he observes when the candle flame is gradually moved away from the pole of the mirror. Draw a ray diagram to show the image formation when the object distance is nearly equal to the radius of curvature of the mirror.
(ii) A convex mirror used for rear-view on an automobile has a focal length of 3.0 m. If a bus is located at 6.0 m from this mirror, use mirror formula to find the position of the image of the bus as seen in the mirror.
(ii) An object is placed at a distance of 30 cm from the optical centre of a concave lens of focal length 20 cm. Use Lens formula to determine the position of the image formed in this case.
OR
(b) (i) A student wishes to study the image formation by a concave mirror using candle flame as object. State the type of the image formed by the mirror and mention the change in the image formed, if any, that he observes when the candle flame is gradually moved away from the pole of the mirror. Draw a ray diagram to show the image formation when the object distance is nearly equal to the radius of curvature of the mirror.
(ii) A convex mirror used for rear-view on an automobile has a focal length of 3.0 m. If a bus is located at 6.0 m from this mirror, use mirror formula to find the position of the image of the bus as seen in the mirror.
(a) (i) Glass Slab Refraction:
Refraction happens twice: Air to Glass (Denser) and Glass to Air (Rarer).
Since the refracting surfaces are parallel, the bending at the first surface is exactly reversed at the second surface. Hence, the emergent ray is parallel to the incident ray but laterally displaced.
Normal Incidence: No deviation occurs. The ray passes straight.
(a) (ii) Concave Lens:
Given: \(u = -30\) cm, \(f = -20\) cm (Concave).
Lens Formula: \(\frac{1}{v} - \frac{1}{u} = \frac{1}{f}\)
\(\frac{1}{v} = \frac{1}{f} + \frac{1}{u}\)
\(\frac{1}{v} = \frac{1}{-20} + \frac{1}{-30} = -\frac{1}{20} - \frac{1}{30}\)
\(\frac{1}{v} = \frac{-3 - 2}{60} = -\frac{5}{60} = -\frac{1}{12}\)
\(v = -12\) cm.
Image is formed 12 cm on the same side as the object (Virtual).
(b) (i) Concave Mirror Study:
Type of Image (General): Real and Inverted (except when ultra-close, P and F).
Change: As object moves away from Pole (from F to Infinity), the image moves from Infinity towards Focus and its size decreases (from Highly Magnified to Point Size).
Ray Diagram (at C): Object distance \(\approx\) Radius of Curvature.
Meaning image is formed at C, Real, Inverted, Same Size.
(b) (ii) Convex Mirror Rear-View:
Given: \(f = +3.0\) m, \(u = -6.0\) m.
Mirror Formula: \(\frac{1}{v} + \frac{1}{u} = \frac{1}{f}\)
\(\frac{1}{v} = \frac{1}{3} - \frac{1}{-6} = \frac{1}{3} + \frac{1}{6} = \frac{2+1}{6} = \frac{3}{6} = \frac{1}{2}\)
\(v = +2.0\) m.
Image is formed 2.0 m behind the mirror.
Refraction happens twice: Air to Glass (Denser) and Glass to Air (Rarer).
Since the refracting surfaces are parallel, the bending at the first surface is exactly reversed at the second surface. Hence, the emergent ray is parallel to the incident ray but laterally displaced.
Image: Glass Slab Refraction
Normal Incidence: No deviation occurs. The ray passes straight.
Image: Normal Incidence Slab
(a) (ii) Concave Lens:
Given: \(u = -30\) cm, \(f = -20\) cm (Concave).
Lens Formula: \(\frac{1}{v} - \frac{1}{u} = \frac{1}{f}\)
\(\frac{1}{v} = \frac{1}{f} + \frac{1}{u}\)
\(\frac{1}{v} = \frac{1}{-20} + \frac{1}{-30} = -\frac{1}{20} - \frac{1}{30}\)
\(\frac{1}{v} = \frac{-3 - 2}{60} = -\frac{5}{60} = -\frac{1}{12}\)
\(v = -12\) cm.
Image is formed 12 cm on the same side as the object (Virtual).
(b) (i) Concave Mirror Study:
Type of Image (General): Real and Inverted (except when ultra-close, P and F).
Change: As object moves away from Pole (from F to Infinity), the image moves from Infinity towards Focus and its size decreases (from Highly Magnified to Point Size).
Ray Diagram (at C): Object distance \(\approx\) Radius of Curvature.
Image: Object at C Concave Mirror
Meaning image is formed at C, Real, Inverted, Same Size.
(b) (ii) Convex Mirror Rear-View:
Given: \(f = +3.0\) m, \(u = -6.0\) m.
Mirror Formula: \(\frac{1}{v} + \frac{1}{u} = \frac{1}{f}\)
\(\frac{1}{v} = \frac{1}{3} - \frac{1}{-6} = \frac{1}{3} + \frac{1}{6} = \frac{2+1}{6} = \frac{3}{6} = \frac{1}{2}\)
\(v = +2.0\) m.
Image is formed 2.0 m behind the mirror.
1 Mark
Q4. An optical device 'X' is placed obliquely in
the path of a narrow parallel beam of light. If the emergent beam gets displaced laterally, the device
'X' is :
Lateral Displacement: Occurs when light passes through a
parallel-sided transparent medium.
A glass slab causes lateral displacement.
Correct Option: (c)
A glass slab causes lateral displacement.
Correct Option: (c)
1 Mark
Q16. To get an image of magnification (-1) on a
screen using a lens of focal length 20 cm, the object distance must be :
Magnification m = -1: Real, Inverted, Same size.
This happens at (2F).
Given (f = 20 \text{ cm}).
Object distance (u = 2f = 2 \times 20 = 40 \text{ cm}).
Correct Option: (c)
This happens at (2F).
Given (f = 20 \text{ cm}).
Object distance (u = 2f = 2 \times 20 = 40 \text{ cm}).
Correct Option: (c)
3 Marks
Q27. If we want to obtain a virtual and magnified
image of an object by using a concave mirror of focal length 18 cm, where should the object be placed?
Use mirror formula to determine the object distance for an image of magnification +2 produced by this
mirror to justify your answer.
Placement: Between Pole (P) and Focus (F). i.e., distance < 18
cm.
Calculation:
\(m = -\frac{v}{u} = +2 \Rightarrow v = -2u\).
Mirror Formula: \(\frac{1}{v} + \frac{1}{u} = \frac{1}{f}\)
Substitute \(v = -2u\): \(\frac{1}{-2u} + \frac{1}{u} = \frac{-1 + 2}{2u} = \frac{1}{2u}\).
So, \(\frac{1}{2u} = \frac{1}{-18}\) (Concave f is negative).
\(2u = -18 \Rightarrow u = -9 \text{ cm}\).
Object is placed at 9 cm from the mirror (which is < 18 cm).
Calculation:
\(m = -\frac{v}{u} = +2 \Rightarrow v = -2u\).
Mirror Formula: \(\frac{1}{v} + \frac{1}{u} = \frac{1}{f}\)
Substitute \(v = -2u\): \(\frac{1}{-2u} + \frac{1}{u} = \frac{-1 + 2}{2u} = \frac{1}{2u}\).
So, \(\frac{1}{2u} = \frac{1}{-18}\) (Concave f is negative).
\(2u = -18 \Rightarrow u = -9 \text{ cm}\).
Object is placed at 9 cm from the mirror (which is < 18 cm).
1 Mark
Q16. In order to obtain large images of the
teeth of patients, the dentist holds the concave mirror in such a manner that the teeth are
positioned
Dentist Mirror (Concave):
To see an erect, virtual, and magnified image, the object (teeth) must be placed between the Pole (P) and Focus (F).
Correct Option: (B)
To see an erect, virtual, and magnified image, the object (teeth) must be placed between the Pole (P) and Focus (F).
Correct Option: (B)
2 Marks
Q21. During an experiment, a student observes
that the given mirror is forming an erect and diminished image for all positions of the object
placed in front of it.
Based on this information, answer the following questions :
(a) Identify the mirror.
(b) Using new Cartesian sign convention, write the magnitude of the magnification in this case.
(c) What will be the radius of curvature of the mirror if its focal length is 20 cm ?
Based on this information, answer the following questions :
(a) Identify the mirror.
(b) Using new Cartesian sign convention, write the magnitude of the magnification in this case.
(c) What will be the radius of curvature of the mirror if its focal length is 20 cm ?
(a) Convex Mirror (Always forms erect and diminished image).
(b) Magnification (m) is positive (erect) and less than 1 (diminished). So, 0 < m < 1.
(c) Radius of curvature \(R = 2f\).
Given \(f = +20\text{ cm}\) (Convex mirror focal length is positive).
\(R = 2 \times 20 = 40\text{ cm}\).
(b) Magnification (m) is positive (erect) and less than 1 (diminished). So, 0 < m < 1.
(c) Radius of curvature \(R = 2f\).
Given \(f = +20\text{ cm}\) (Convex mirror focal length is positive).
\(R = 2 \times 20 = 40\text{ cm}\).
3 Marks
Q29. The power of a lens is \(-0.25\text{
D}\). Based on this information, find out
(a) The type of lens and its focal length.
(b) The eye defect for which it may be used as a corrective lens.
(c) The nature and size of the image formed by this lens when an object is placed between F and 2F from the optical centre of this lens.
(a) The type of lens and its focal length.
(b) The eye defect for which it may be used as a corrective lens.
(c) The nature and size of the image formed by this lens when an object is placed between F and 2F from the optical centre of this lens.
(a) Power \(P = -0.25\text{ D}\). Negative sign indicates Concave Lens.
Focal length \(f = \frac{1}{P} = \frac{1}{-0.25} = -4\text{ m}\) or \(-400\text{ cm}\).
(b) Concave lens is used to correct Myopia (Near-sightedness).
(c) For a Concave lens, regardless of object position (between F and 2F or anywhere), the image is always Virtual, Erect and Diminished.
Focal length \(f = \frac{1}{P} = \frac{1}{-0.25} = -4\text{ m}\) or \(-400\text{ cm}\).
(b) Concave lens is used to correct Myopia (Near-sightedness).
(c) For a Concave lens, regardless of object position (between F and 2F or anywhere), the image is always Virtual, Erect and Diminished.
5 Marks
Q36. (A) (a) Observe the following diagram
and compare (i) speed of light and (ii) optical densities of the three media A, B and C. Also give
justification for your answer of any one of the two cases in terms of refractive indices of A, B and
C.
(b) Redraw the path of a ray of light through the three media, if the ray of light starting from medium A falls on the medium B
(i) Obliquely and the optical density of medium B is made more than that of A and C.
(ii) The ray falls normally from medium A to medium B.
OR
(B) Analyse the following observation table showing variation of image distance (v) with object distance (u) in case of a convex lens and answer the questions that follow without doing any calculations :
[Table logic: u=-15, v=-60; u=-25, v=+100; u=-30, v=+60; u=-40, v=+40; u=-60, v=+30; u=-100, v=+25]
(a) Determine the focal length of the lens. Give reason for your answer.
(b) Find magnification of the image formed in Observation No. 3.
(c) The numerical value of magnifications in cases of observation 1 and 2 is same. List two differences in the images formed in these two cases.
(b) Redraw the path of a ray of light through the three media, if the ray of light starting from medium A falls on the medium B
(i) Obliquely and the optical density of medium B is made more than that of A and C.
(ii) The ray falls normally from medium A to medium B.
OR
(B) Analyse the following observation table showing variation of image distance (v) with object distance (u) in case of a convex lens and answer the questions that follow without doing any calculations :
[Table logic: u=-15, v=-60; u=-25, v=+100; u=-30, v=+60; u=-40, v=+40; u=-60, v=+30; u=-100, v=+25]
(a) Determine the focal length of the lens. Give reason for your answer.
(b) Find magnification of the image formed in Observation No. 3.
(c) The numerical value of magnifications in cases of observation 1 and 2 is same. List two differences in the images formed in these two cases.
(A)
(a) From diagram: Ray bends toward normal from A to B (B is denser than A). Ray bends away from normal from B to C (B is denser than C). Ray direction C implies it is parallel to A?
Comparing Densities: \(n_B > n_A\) and \(n_B > n_C\).
Speed: \(v \propto 1/n\). So Speed in A > Speed in B. Speed in C > Speed in B.
(b) Placeholders for Ray Diagrams:
OR
(B)
(a) At Observation 4: \(u = -40, v = +40\). Since \(|u| = |v| = 2f\), this is the center of curvature.
\(2f = 40 \Rightarrow f = +20\text{ cm}\).
(b) Obs 3: \(u = -30, v = +60\).
\(m = v/u = 60 / -30 = -2\).
(c) Obs 1 (u=-15, v=-60): Virtual image (u < f). m is Positive (+4).
Obs 2 (u=-25, v=+100): Real image. m is Negative (-4).
Note: Question says "numerical value... is same". |+4| = |-4| = 4. True.
Differences:
1. Obs 1 image is Virtual & Erect. Obs 2 image is Real & Inverted.
2. Obs 1 formed on same side. Obs 2 formed on other side.
(a) From diagram: Ray bends toward normal from A to B (B is denser than A). Ray bends away from normal from B to C (B is denser than C). Ray direction C implies it is parallel to A?
Comparing Densities: \(n_B > n_A\) and \(n_B > n_C\).
Speed: \(v \propto 1/n\). So Speed in A > Speed in B. Speed in C > Speed in B.
(b) Placeholders for Ray Diagrams:
OR
(B)
(a) At Observation 4: \(u = -40, v = +40\). Since \(|u| = |v| = 2f\), this is the center of curvature.
\(2f = 40 \Rightarrow f = +20\text{ cm}\).
(b) Obs 3: \(u = -30, v = +60\).
\(m = v/u = 60 / -30 = -2\).
(c) Obs 1 (u=-15, v=-60): Virtual image (u < f). m is Positive (+4).
Obs 2 (u=-25, v=+100): Real image. m is Negative (-4).
Note: Question says "numerical value... is same". |+4| = |-4| = 4. True.
Differences:
1. Obs 1 image is Virtual & Erect. Obs 2 image is Real & Inverted.
2. Obs 1 formed on same side. Obs 2 formed on other side.
1 Mark
Q14. Identify from the following ray diagram
which shows the correct path of the reflected ray for the ray incident on a concave mirror as shown
:
The incident ray is parallel to the principal axis.
Rule: A ray parallel to the principal axis, after reflection from a concave mirror, passes through the Principal Focus (F).
Figure (D) typically shows this path correctly.
Correct Option: (D) (Assuming D shows passing through F).
Rule: A ray parallel to the principal axis, after reflection from a concave mirror, passes through the Principal Focus (F).
Figure (D) typically shows this path correctly.
Correct Option: (D) (Assuming D shows passing through F).
2 Marks
Q22. Study the figure in which the path of a
ray of light going from Medium 1 to Medium 2 is shown.
(a) Out of the two Media - Medium 1 and Medium 2, in which is the speed of light more ?
(b) State reason of bending of the refracted ray away from the normal.
(c) Express refractive index of Medium 2 with respect to Medium 1 in terms of speed of light in two media.
(a) Out of the two Media - Medium 1 and Medium 2, in which is the speed of light more ?
(b) State reason of bending of the refracted ray away from the normal.
(c) Express refractive index of Medium 2 with respect to Medium 1 in terms of speed of light in two media.
(a) Ray bends away from normal (Angle r > Angle i). This means light travels
from Denser to Rarer. Speed is more in Medium 2 (Rarer).
(b) Change in speed of light. Speed increases in rarer medium causing it to bend away.
(c) \(n_{21} = \frac{v_1}{v_2}\) (Speed in Med 1 / Speed in Med 2).
(b) Change in speed of light. Speed increases in rarer medium causing it to bend away.
(c) \(n_{21} = \frac{v_1}{v_2}\) (Speed in Med 1 / Speed in Med 2).
3 Marks
Q29. A student placed a candle flame at
different distances from a convex lens and focused its image on a screen. He recorded his
observation in tabular form...
[Table Data: 1. u=-90 v=+18; 2. u=-60 v=+20; 3. u=-40 v=+24; 4. u=-30 v=+30; 5. u=-24 v=+40; 6. u=-20 v=+60; 7. u=-18 v=+90; 8. u=-12 v=+120]
(a) What is the focal length of the convex lens used ? Give reason.
(b) Which one of the sets of observations is not correct and why ?
(c) Draw ray diagram to show image formation for any correct set of observation.
[Table Data: 1. u=-90 v=+18; 2. u=-60 v=+20; 3. u=-40 v=+24; 4. u=-30 v=+30; 5. u=-24 v=+40; 6. u=-20 v=+60; 7. u=-18 v=+90; 8. u=-12 v=+120]
(a) What is the focal length of the convex lens used ? Give reason.
(b) Which one of the sets of observations is not correct and why ?
(c) Draw ray diagram to show image formation for any correct set of observation.
(a) From Obs 4: \(|u| = 30, |v| = 30\). Since \(u=v=2f\), \(2f = 30\), so \(f = 15\text{
cm}\).
(b) Observation 8 (u=-12, v=+120).
Reason: For \(u = 12\) cm (which is < f=15 cm), image should be Virtual (same side, negative v). But v is +120 (Real). Hence incorrect.
(c)
(b) Observation 8 (u=-12, v=+120).
Reason: For \(u = 12\) cm (which is < f=15 cm), image should be Virtual (same side, negative v). But v is +120 (Real). Hence incorrect.
(c)
Ray Diagram for Convex Lens
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