Vardaan Institute - Physics Solutions (Level 0)

Topic 1: Reflection & Spherical Mirrors

Part I: Fill in the Blanks

The angle of incidence is always equal to the angle of ________.

Solution: Reflection ($\angle i = \angle r$).

The formula relating focal length ($f$) and radius of curvature ($R$) of a spherical mirror is $f =$ ________.

Solution: $R/2$.

The mirror formula connecting object distance ($u$), image distance ($v$), and focal length ($f$) is ________.

Solution: $\frac{1}{f} = \frac{1}{v} + \frac{1}{u}$.

The linear magnification ($m$) produced by a spherical mirror is given by the ratio of the height of the image to the height of the ________.

Solution: Object ($h'/h$).

In terms of $v$ and $u$, the magnification formula for a mirror is $m =$ ________.

Solution: $-v/u$.

Part II: Multiple Choice Questions (MCQs)

The focal length of a plane mirror is:

Solution: (b) Infinite (It is a sphere of infinite radius).

A convex mirror always produces an image which is:

Solution: (c) Virtual, erect, diminished.

AI Prompt: Create a mathematically correct and highly accurate physics ray diagram showing image formation by a convex mirror. Show a real object, principal axis, focal point (F), center of curvature (C), and the two standard diverging rays producing a virtual, erect, diminished image behind the mirror. The background of the whole image should be fully white. It should be in landscape mode. High quality, best resolution, clean precise lines.

File Name: Level0_Q7_ConvexMirrorRayDiagram.png

If magnification $m = -2$ for a mirror, the negative sign indicates that the image is:

Solution: (b) Inverted (Real images are inverted and have negative magnification).

Which mirror is commonly used as a rear-view mirror in vehicles?

Solution: (c) Convex mirror (Because it provides a wider field of view and always forms an erect image).

Part III: Basic Numericals & Concepts

10.

The radius of curvature of a convex mirror is $20\text{ cm}$. Find its focal length.

Solution: $f = R/2 = +20 / 2 = +10\text{ cm}$. (Positive because the focus of a convex mirror lies behind it).

11.

An object is placed $15\text{ cm}$ in front of a concave mirror of focal length $10\text{ cm}$. Calculate the image distance ($v$).

Solution: Using mirror formula: $\frac{1}{f} = \frac{1}{v} + \frac{1}{u}$. Here, $u = -15\text{ cm}$, $f = -10\text{ cm}$.
$\frac{1}{v} = \frac{1}{-10} - \frac{1}{-15} = \frac{-1}{10} + \frac{1}{15} = \frac{-3 + 2}{30} = \frac{-1}{30}$.
Therefore, $v = -30\text{ cm}$. (The image is formed $30\text{ cm}$ in front of the mirror).

12.

Define the Principal Axis of a spherical mirror.

Solution: The straight line passing through the pole ($P$) and the center of curvature ($C$) of a spherical mirror is called its principal axis.

13.

The magnification produced by a mirror is $+3$. What does this mean?

Solution: The positive sign indicates that the image is virtual and erect. The number '3' indicates that the image is magnified (3 times larger than the object).

Topic 2: Refraction & Total Internal Reflection (TIR)

Part I: Fill in the Blanks

14.

The ratio of the sine of the angle of incidence to the sine of the angle of refraction is a constant called the ________.

Solution: Refractive Index (Snell's Law).

15.

The absolute refractive index ($n$) of a medium is given by $n = c /$ ________.

Solution: $v$ (Speed of light in the medium).

16.

When a light ray travels from a denser to a rarer medium, it bends ________ from the normal.

Solution: Away.

17.

The formula relating real depth ($h$) and apparent depth ($h'$) with refractive index ($n$) is $n =$ ________.

Solution: Real Depth / Apparent Depth ($h / h'$).

18.

For Total Internal Reflection to occur, the angle of incidence must be greater than the ________ angle.

Solution: Critical.

Part II: Multiple Choice Questions (MCQs)

19.

The fundamental cause of refraction is the change in ________ of light when it enters a different medium.

Solution: (b) Speed.

20.

Which phenomenon is responsible for the sparkling of a diamond?

Solution: (c) Total Internal Reflection (Diamond has a very high refractive index and low critical angle).

21.

Optical fibers work on the principle of:

Solution: (b) Total Internal Reflection.

AI Prompt: Create a mathematically correct and highly accurate physics ray diagram showing Total Internal Reflection (TIR) inside a cylindrical optical fiber core. Show the core-cladding boundary, a light ray entering, and bouncing off the internal walls (TIR) as it travels through the fiber. The background of the whole image should be fully white. It should be in landscape mode. High quality, best resolution, clean precise lines.

File Name: Level0_Q21_OpticalFiberTIR.png

22.

Mirage in deserts is an optical illusion caused by:

Solution: (c) TIR due to atmospheric temperature gradients (Hot air near sand is rarer, cold air above is denser).

Part III: Basic Numericals & Concepts

23.

The speed of light in a vacuum is $3 \times 10^8 \text{ m/s}$. Find the speed of light in glass of refractive index $1.5$.

Solution: Refractive index $n = c / v \Rightarrow v = c / n$.
$v = \frac{3 \times 10^8}{1.5} = 2 \times 10^8 \text{ m/s}$.

24.

The refractive index of water is $4/3$. Write the expression for its critical angle ($C$) with respect to air.

Solution: $\sin C = \frac{1}{n} = \frac{1}{4/3} = \frac{3}{4}$. Therefore, $C = \sin^{-1}(0.75)$.

25.

A coin is placed at the bottom of a beaker containing water to a depth of $12\text{ cm}$. By what depth does the coin appear to be raised? (Refractive index of water $= 4/3$).

Solution: Apparent depth $h' = \frac{\text{Real Depth}}{n} = \frac{12}{4/3} = \frac{12 \times 3}{4} = 9\text{ cm}$.
The coin appears raised by = Real Depth - Apparent Depth = $12 - 9 = 3\text{ cm}$.

26.

State the two necessary conditions for Total Internal Reflection to occur.

Solution: (1) Light must travel from an optically denser medium to an optically rarer medium.
(2) The angle of incidence in the denser medium must be greater than the critical angle for the given pair of media ($i > C$).

Topic 3: Lenses & Power

Part I: Fill in the Blanks

27.

The thin lens formula connecting $v$, $u$, and $f$ is ________.

Solution: $\frac{1}{f} = \frac{1}{v} - \frac{1}{u}$.

28.

Power of a lens ($P$) is the reciprocal of its ________.

Solution: Focal length ($f$ in meters).

29.

The SI unit of power of a lens is ________.

Solution: Diopter (D).

30.

If two thin lenses of power $P_1$ and $P_2$ are placed in contact, the equivalent power $P =$ ________.

Solution: $P_1 + P_2$.

31.

A convex lens is thicker at the ________ and thinner at the edges.

Solution: Center (Middle).

Part II: Multiple Choice Questions (MCQs)

32.

A convex lens of focal length $f$ forms a real image of the same size as the object. The object is placed at:

Solution: (d) $2F$ (Distance of $2f$ from the optical center).

33.

The power of a concave lens is always:

Solution: (b) Negative (Because its focal length is negative).

34.

If the focal length of a lens is $0.5\text{ m}$, its power is:

Solution: (b) $+2.0\text{ D}$ ($P = 1/f = 1/0.5 = +2$).

35.

Lens Maker's formula states that $1/f = (n - 1)($________ $)$.

Solution: (b) $\frac{1}{R_1} - \frac{1}{R_2}$.

Part III: Basic Numericals & Concepts

36.

The focal length of a convex lens is $25\text{ cm}$. Find its power.

Solution: Focal length in meters $f = +0.25\text{ m}$.
Power $P = 1/f = 1/0.25 = +4.0\text{ D}$.

37.

Two thin lenses of power $+5\text{ D}$ and $-2\text{ D}$ are placed in contact. Find the equivalent power and the focal length of the combination.

Solution: Equivalent Power $P = P_1 + P_2 = (+5) + (-2) = +3\text{ D}$.
Focal length $F = 1/P = 1/3\text{ m} = +33.3\text{ cm}$.

38.

An object is placed $20\text{ cm}$ in front of a convex lens of focal length $10\text{ cm}$. Find the image distance.

Solution: Lens formula: $\frac{1}{f} = \frac{1}{v} - \frac{1}{u}$. Here, $u = -20\text{ cm}$, $f = +10\text{ cm}$.
$\frac{1}{10} = \frac{1}{v} - \frac{1}{-20} \Rightarrow \frac{1}{10} = \frac{1}{v} + \frac{1}{20}$.
$\frac{1}{v} = \frac{1}{10} - \frac{1}{20} = \frac{2-1}{20} = \frac{1}{20}$.
Thus, $v = +20\text{ cm}$ (Real image formed on the other side).

39.

What is the power of a plane glass plate?

Solution: Zero. (A plane glass plate has infinite focal length, so $P = 1/\infty = 0$).

Topic 4: Prism & Dispersion

Part I: Fill in the Blanks

40.

The formula connecting angle of prism ($A$), angle of deviation ($\delta$), angle of incidence ($i$) and angle of emergence ($e$) is $A + \delta =$ ________.

Solution: $i + e$.

41.

The phenomenon of splitting of white light into its constituent colors is called ________.

Solution: Dispersion.

42.

For an equilateral glass prism, the angle of the prism ($A$) is ________ degrees.

Solution: $60^\circ$.

43.

At the position of minimum deviation, the angle of incidence ($i$) is ________ to the angle of emergence ($e$).

Solution: Equal ($i = e$).

Part II: Multiple Choice Questions (MCQs)

44.

When white light passes through a glass prism, which color deviates the most?

Solution: (d) Violet (It has the shortest wavelength and highest refractive index).

AI Prompt: Create a mathematically correct and highly accurate physics ray diagram showing dispersion of white light through a triangular glass prism. Show a single white beam entering, refracting, and splitting into a spectrum of colors (Red at the top, Violet at the bottom) upon emerging. Label the prism angle A and angle of deviation delta. The background of the whole image should be fully white. It should be in landscape mode. High quality, best resolution, clean precise lines.

File Name: Level0_Q44_PrismDispersion.png

45.

Which color has the maximum speed in a glass prism?

Solution: (c) Red (Speed $v = c/n$. Red has lowest $n$, hence maximum speed inside glass).

46.

The relation $\mu = \frac{\sin((A+\delta_m)/2)}{\sin(A/2)}$ is used to find the refractive index of a:

Solution: (c) Prism.

Part III: Basic Numericals & Concepts

47.

A glass prism has an angle of $60^\circ$ and produces a minimum deviation of $30^\circ$. Calculate the refractive index of the glass material. ($\sin 45^\circ = 1/\sqrt{2}, \sin 30^\circ = 1/2$)

Solution: Using prism formula: $\mu = \frac{\sin(\frac{A+\delta_m}{2})}{\sin(\frac{A}{2})}$.
Here $A = 60^\circ$, $\delta_m = 30^\circ$.
$\mu = \frac{\sin(\frac{60+30}{2})}{\sin(\frac{60}{2})} = \frac{\sin 45^\circ}{\sin 30^\circ} = \frac{1/\sqrt{2}}{1/2} = \frac{2}{\sqrt{2}} = \sqrt{2} \approx 1.414$.

48.

Define the 'Angle of Deviation' in a prism.

Solution: It is the angle between the extended incident ray and the emergent ray. It measures how much the light ray has bent from its original path after passing through the prism.

49.

Which color of light has the longest wavelength in the visible spectrum?

Solution: Red light.

Topic 5: Optical Instruments

Part I: Fill in the Blanks

50.

A simple microscope consists of a single ________ lens.

Solution: Convex (of short focal length).

51.

In a compound microscope, the lens near the object is called the ________ lens.

Solution: Objective.

52.

In a compound microscope, the focal length of the objective lens is ________ than the focal length of the eyepiece.

Solution: Less (Shorter).

53.

Astronomical telescopes are used to observe very ________ objects like stars and planets.

Solution: Distant (Far away).

54.

In a reflecting telescope, a ________ mirror is used instead of an objective lens.

Solution: Concave (or Parabolic).

Part II: Multiple Choice Questions (MCQs)

55.

The magnifying power ($M$) of an astronomical telescope in normal adjustment is given by:

Solution: (b) $f_o / f_e$ (Where $f_o$ is focal length of objective and $f_e$ is focal length of eyepiece).

AI Prompt: Create a mathematically correct and highly accurate physics ray diagram of an Astronomical Telescope in normal adjustment (final image at infinity). Show a large objective lens, a small eyepiece lens, principal axis, parallel rays from infinity entering the objective, forming a real inverted image at the common focal point, and exiting the eyepiece as parallel rays. The background of the whole image should be fully white. It should be in landscape mode. High quality, best resolution, clean precise lines.

File Name: Level0_Q55_AstronomicalTelescope.png

56.

What is a major advantage of a reflecting telescope over a refracting telescope?

Solution: (b) It is free from chromatic aberration (Mirrors reflect all colors identically).

57.

The final image formed by a standard compound microscope is:

Solution: (b) Virtual and inverted (with respect to original object).

Part III: Basic Numericals & Concepts

58.

Write the formula for the magnifying power of a simple microscope when the final image is formed at the least distance of distinct vision ($D$).

Solution: $M = 1 + \frac{D}{f}$. (Where $D = 25\text{ cm}$ for a normal eye).

59.

An astronomical telescope has an objective lens of focal length $100\text{ cm}$ and an eyepiece of focal length $5\text{ cm}$. Find its magnifying power in normal adjustment.

Solution: $M = \frac{f_o}{f_e} = \frac{100}{5} = 20$.

60.

For the telescope in the previous question, calculate the length of the telescope tube in normal adjustment.

Solution: In normal adjustment, the length of the tube $L = f_o + f_e = 100 + 5 = 105\text{ cm}$.

61.

Why is the objective lens of an astronomical telescope made with a very large aperture (diameter)?

Solution: A large aperture allows the objective lens to gather more light from faint, distant astronomical objects, producing a brighter and more resolved image.

Section: True/False & Match the Following

Part I: True or False

62.

The speed of light in water is less than the speed of light in a vacuum.

Solution: True (Because water is optically denser than vacuum).

63.

Magnification of a plane mirror is always exactly $+1$.

Solution: True (Image is virtual, erect, and the exact same size).

64.

If two thin lenses of power $+2\text{D}$ and $-1\text{D}$ are combined, the net power is $+3\text{D}$.

Solution: False (Net power $P = P_1 + P_2 = (+2) + (-1) = +1\text{D}$).

65.

Myopia (short-sightedness) is corrected using a convex lens.

Solution: False (It is corrected using a concave/diverging lens).

66.

The critical angle for total internal reflection depends on the refractive index of the medium.

Solution: True ($\sin C = 1/n$).

67.

In an optical fiber, the refractive index of the core is less than that of the cladding.

Solution: False (Core must be denser than cladding for TIR to occur, so core index is higher).

Part II: Match the Following (Formulas & Concepts)

68.

Match the laws and formulas with their mathematical expressions:

Solution:
(P) Snell's Law $\rightarrow$ (3) $\frac{n_2}{n_1} = \frac{\sin i}{\sin r}$
(Q) Lens Maker's Formula $\rightarrow$ (1) $\frac{1}{f} = (n-1)(\frac{1}{R_1} - \frac{1}{R_2})$
(R) Mirror Formula $\rightarrow$ (2) $\frac{1}{f} = \frac{1}{v} + \frac{1}{u}$
(S) Thin Lens Formula $\rightarrow$ (4) $\frac{1}{f} = \frac{1}{v} - \frac{1}{u}$

69.

Match the optical phenomena/instruments with their primary working principle:

Solution:
(P) Optical Fiber $\rightarrow$ (3) Total Internal Reflection (TIR)
(Q) Cassegrain Telescope $\rightarrow$ (4) Reflection via Parabolic/Concave mirror
(R) Rainbow Formation $\rightarrow$ (1) Dispersion of light (along with TIR and Refraction)
(S) Compound Microscope $\rightarrow$ (2) Refraction through two convex lenses

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