Vardaan Learning Institute

Ans: The laws of reflection are:

1. The angle of incidence ($\theta_i$) is always equal to the angle of reflection ($\theta_r$).
2. The incident ray, the normal to the reflecting surface at the point of incidence, and the reflected ray all lie in the same plane.

Yes, these laws hold true for all types of reflecting surfaces (including irregular, rough, and curved surfaces like concave/convex mirrors). The apparent difference in irregular reflection (diffuse reflection) occurs because the normal at different points of a rough surface points in different directions, causing reflected rays to scatter, though the laws are satisfied at every microscopic point.

Ans: The differences are:

1. Real Image: Formed when light rays actually meet after reflection/refraction. It is always inverted and can be obtained on a screen.
2. Virtual Image: Formed when light rays do not actually meet but appear to diverge from a point behind the mirror/lens. It is always erect and cannot be obtained on a screen.

Yes, a virtual image can be photographed because the camera lens converges the diverging rays entering it, forming a real image on the camera sensor/film.

Ans: Spherical Mirror Definitions:

1. Center of Curvature (C): The center of the hollow sphere of glass of which the mirror is a part.
2. Principal Axis: The straight line passing through the pole (P) and the center of curvature (C) of the mirror.
3. Focus (F): The point on the principal axis where parallel rays of light meet (converge) or appear to diverge from after reflection.
4. Focal Length ($f$): The distance between the pole (P) and the focus (F) of the mirror.

For a spherical mirror of small aperture, the focus (F) lies exactly midway between the pole (P) and the center of curvature (C). Hence, the radius of curvature ($R$) is twice the focal length ($f$), represented mathematically as: $$R = 2f$$

Ans: The three rules for image construction in spherical mirrors are:

1. A ray parallel to the principal axis passes through the focus (concave) or appears to diverge from the focus (convex) after reflection.
2. A ray passing through the focus (concave) or directed towards the focus (convex) becomes parallel to the principal axis after reflection.
3. A ray passing through the center of curvature (C) is reflected back along the same path because it falls normally on the mirror surface ($\theta_i = \theta_r = 0^\circ$).

Ans: Concave mirror image properties:

1. At infinity: Image formed at Focus (F); real, inverted, highly diminished (point-sized).
2. Beyond C: Image formed between C and F; real, inverted, diminished.
3. At C: Image formed at C; real, inverted, same size as object.
4. Between C and F: Image formed beyond C; real, inverted, magnified.
5. At F: Image formed at infinity; real, inverted, highly magnified.
6. Between P and F: Image formed behind the mirror; virtual, erect, magnified.

Ans: A convex mirror is used as a rear-view mirror because:

1. It always forms an erect (upright) and diminished image of the traffic behind, making it easy to identify vehicles.
2. It has a much wider field of view because its outwardly curved surface allows it to collect light from a larger angular area compared to a flat plane mirror or a concave mirror.

Ans: New Cartesian Sign Conventions:

1. The object is always placed to the left of the mirror (light travels from left to right).
2. All distances parallel to the principal axis are measured from the pole (P) acting as the origin.
3. Distances measured in the direction of incident light (to the right of P) are positive; distances opposite (to the left of P) are negative.
4. Heights measured upwards and perpendicular to the principal axis are positive; downwards are negative.

A concave mirror's focus lies in front of it (to the left), making its focal length negative ($-f$). A convex mirror's focus lies behind it (to the right), making its focal length positive ($+f$).

Ans: Mirror Formula: $$\frac{1}{f} = \frac{1}{v} + \frac{1}{u}$$ Magnification Formula: $$m = \frac{h_i}{h_o} = -\frac{v}{u}$$ Significance: - (a) A negative magnification ($m < 0$) indicates that the image is real and inverted. - (b) A positive magnification ($m > 0$) indicates that the image is virtual and erect.

Ans: Given: $h_o = +4\text{ cm}$, $u = -25\text{ cm}$, $f = -15\text{ cm}$ (concave mirror).
Using mirror formula: $\frac{1}{v} = \frac{1}{f} - \frac{1}{u}$ $$\frac{1}{v} = \frac{1}{-15} - \frac{1}{-25} = -\frac{1}{15} + \frac{1}{25} = \frac{-5 + 3}{75} = -\frac{2}{75}$$ $$v = -37.5\text{ cm}$$ The image is formed at $37.5\text{ cm}$ in front of the mirror (real image).
Using magnification: $m = -\frac{v}{u} = -\frac{-37.5}{-25} = -1.5$
Height of image: $h_i = m \times h_o = -1.5 \times 4 = -6\text{ cm}$.
Nature: Real, inverted, and magnified ($6\text{ cm}$ tall).

Ans: Given: $R = +3.0\text{ m} \Rightarrow f = +1.5\text{ m}$ (convex mirror), $u = -5.0\text{ m}$.
Using mirror formula: $\frac{1}{v} = \frac{1}{f} - \frac{1}{u}$ $$\frac{1}{v} = \frac{1}{1.5} - \frac{1}{-5.0} = \frac{10}{15} + \frac{1}{5} = \frac{2}{3} + \frac{1}{5} = \frac{10 + 3}{15} = \frac{13}{15}$$ $$v = +\frac{15}{13} \approx +1.15\text{ m}$$ The image is formed at $1.15\text{ m}$ behind the mirror.
Magnification: $m = -\frac{v}{u} = -\frac{1.15}{-5.0} \approx +0.23$.
Nature: Virtual, erect, and diminished to $0.23$ times the size of the bus.

Ans: Given: $u = -10\text{ cm}$, real image $\Rightarrow m = -3$.
Using $m = -\frac{v}{u}$: $$-3 = -\frac{v}{-10} \Rightarrow v = -30\text{ cm}$$ The image is located at $30\text{ cm}$ in front of the mirror.
Using mirror formula to find $f$: $$\frac{1}{f} = \frac{1}{v} + \frac{1}{u} = \frac{1}{-30} + \frac{1}{-10} = \frac{-1 - 3}{30} = -\frac{4}{30} = -\frac{2}{15}$$ $$f = -7.5\text{ cm}$$ The focal length of the concave mirror is $7.5\text{ cm}$.

Ans: The power of a lens ($P$) is a measure of its ability to converge or diverge light rays. It is defined as the reciprocal of its focal length ($f$) in meters: $$P = \frac{1}{f\text{ (in meters)}}$$ SI Unit: Dioptre (D).
One dioptre is the power of a lens whose focal length is exactly $1\text{ meter}$ ($1\text{ D} = 1\text{ m}^{-1}$).

Ans: Given: $P = +2.0\text{ D}$.
Using $P = \frac{1}{f}$: $$f = \frac{1}{P} = \frac{1}{+2.0} = +0.5\text{ m} = +50\text{ cm}$$ The focal length of the lens is $+50\text{ cm}$. Since the focal length and power are positive, it is a converging (convex) lens.

Ans: Given: $f = -2\text{ m}$ (concave lens focal length is negative).
Using $P = \frac{1}{f}$: $$P = \frac{1}{-2} = -0.5\text{ D}$$ The power of the concave lens is $-0.5\text{ Dioptres}$ (negative optical sign).

Ans: For two thin lenses in contact:
Net Power: $P = P_1 + P_2$
Net Focal Length: $\frac{1}{f} = \frac{1}{f_1} + \frac{1}{f_2}$
Advantage: Using a lens combination helps minimize optical aberrations (such as chromatic aberration), improves image sharpness, and allows variable zoom/magnification control in devices like cameras, microscopes, and telescopes.

Ans:

1. Dr. Mehta uses a concave mirror.
2. The tooth must be placed between the pole (P) and the focus (F) of the concave mirror. The ray diagram shows diverging rays reflecting from the mirror and being projected back to form a virtual, erect, and highly magnified image behind the mirror.
3. If she pulls the mirror far away, the tooth passes beyond the focus (F) and center of curvature (C). The image will immediately become real and inverted, appearing upside-down and smaller, making the checkup impossible.

Ans:

1. A concave mirror is used. The cooking pot should be placed exactly at the principal focus (F) of the mirror.
2. Parallel sunlight rays falling on the concave surface reflect and meet (converge) at the focus (F), concentrating all the thermal energy at one spot.
3. Black surfaces are excellent absorbers of radiant heat, maximizing thermal capture and heating the pot quickly.

Ans:

1. The optical phenomenon is refraction of light.
2. Refractive index of water: $$n_w = \frac{\text{Real Depth}}{\text{Apparent Depth}} = \frac{3.0\text{ m}}{2.25\text{ m}} = 1.33 \approx \frac{4}{3}$$
3. The ray diagram shows two rays starting from a point bottom O. They travel vertically and obliquely. At the water surface, they bend away from the normal. When projected back, they intersect at a shallower point I, showing the apparent pool bottom.

Ans:

1. Arun is prescribed a concave (diverging) lens.
   Focal length: $f = \frac{1}{P} = \frac{1}{-1.5} \approx -0.67\text{ m} = -67\text{ cm}$.
2. A concave lens is thin at the center and thick at the edges.
3. The ray diagram shows parallel light rays passing through the concave lens and diverging outward, with their virtual projections meeting at the focus ($F$) on the same side.

Ans:

1. Power calculations:
   • Lens A: $f_A = +20\text{ cm} = +0.2\text{ m} \Rightarrow P_A = \frac{1}{+0.2} = +5.0\text{ D}$.
   • Lens B: $f_B = -25\text{ cm} = -0.25\text{ m} \Rightarrow P_B = \frac{1}{-0.25} = -4.0\text{ D}$.
2. Net Power: $P = P_A + P_B = +5.0\text{ D} - 4.0\text{ D} = \mathbf{+1.0\text{ D}}$.
   Net Focal Length: $f = \frac{1}{P} = \frac{1}{+1.0} = \mathbf{+1.0\text{ m}} = \mathbf{+100\text{ cm}}$.
3. Since the net power and net focal length are positive, the combination behaves as a converging (convex) lens system.

Ans:

1. Using Snell's Law for medium A to M: $$n_{MA} = \frac{\sin i}{\sin r} = \frac{\sin 45^\circ}{\sin 30^\circ} = \frac{0.707}{0.5} = 1.414$$ Absolute refractive index of M ($n_M$): $$n_{MA} = \frac{n_M}{n_A} \Rightarrow 1.414 = \frac{n_M}{1.2} \Rightarrow n_M = 1.414 \times 1.2 \approx \mathbf{1.70}$$
2. Speed of light in M: $$v_M = \frac{c}{n_M} = \frac{3 \times 10^8}{1.70} \approx \mathbf{1.76 \times 10^8\text{ m/s}}$$
3. Refractive index of M ($1.70$) is greater than water ($1.33$). Therefore, light enters from a denser medium 'M' into a rarer medium (water) and will bend away from the normal.

Ans: The principal focus of a concave mirror is the point on the principal axis where all incident rays parallel to the axis meet after reflection.
Because a concave mirror is curved inward, the normal to the surface at different points points inward, converging reflected rays. A convex mirror is curved outward, so its normals point outward, causing reflected rays to diverge.

Ans: The face must be held within the focal length (between the pole P and focus F) of the concave mirror. This produces a virtual, erect, and magnified image, enabling the user to see tiny facial hairs clearly.

Ans: Magnification $m = \frac{h_i}{h_o} = +1$. - The number 1 indicates that the size of the image is exactly equal to the size of the object. - The plus sign indicates that the image is virtual and erect.

Ans: No, a convex mirror cannot form a real image of a real object. Because its reflecting surface curves outward, light rays always diverge after reflection and can only be projected back to form a virtual focus behind the mirror, yielding a virtual, erect, and diminished image under all positions.

Ans: Relative refractive index $n_{21} = \frac{v_1}{v_2}$ and $n_{12} = \frac{v_2}{v_1}$.
Multiplying both: $$n_{21} \times n_{12} = \frac{v_1}{v_2} \times \frac{v_2}{v_1} = 1$$ This shows that the relative refractive indices of two media are reciprocals of each other ($n_{21} = \frac{1}{n_{12}}$).

Ans: Snell's Law states: $\frac{\sin i}{\sin r} = \frac{n_2}{n_1}$.
If a ray falls normally, the angle of incidence $i = 0^\circ$.
Using Snell's Law: $$\sin r = \frac{n_1}{n_2} \times \sin 0^\circ = 0 \Rightarrow r = 0^\circ$$ Since the angle of refraction $r = 0^\circ$, the light ray travels straight through without bending.

Ans: Given: Real Depth $= 12\text{ cm}$, $n = \frac{4}{3}$.
Using $n = \frac{\text{Real Depth}}{\text{Apparent Depth}}$: $$\text{Apparent Depth} = \frac{\text{Real Depth}}{n} = \frac{12}{4/3} = 9\text{ cm}$$ Apparent Rise: Real Depth - Apparent Depth $= 12 - 9 = \mathbf{3\text{ cm}}$.
The coin appears to rise by $3\text{ cm}$.

Ans: Lateral displacement increases with slab width because the refracted ray spends more distance traveling obliquely inside the glass, increasing the offset distance before exiting.

Ans: For a convex lens, the image is equal to the object size when placed at $2F_1$.
Given: $v = +50\text{ cm} \Rightarrow 2F_2 = 50\text{ cm} \Rightarrow f = +25\text{ cm} = +0.25\text{ m}$.
The object must be placed at $50\text{ cm}$ in front of the lens ($u = -50\text{ cm}$).
Power of the lens: $$P = \frac{1}{f\text{ (in meters)}} = \frac{1}{+0.25} = \mathbf{+4.0\text{ D}}$$

Ans: The principal focus of a concave lens is called a virtual focus because parallel rays do not actually intersect after passing through the lens. Instead, they diverge, and only their back projections meet at the focus $F_1$, forming a virtual focal point.

Ans: Given: $u = -10\text{ cm}$, $f = +12\text{ cm}$ (convex lens).
Using lens formula: $\frac{1}{v} = \frac{1}{f} + \frac{1}{u}$ $$\frac{1}{v} = \frac{1}{12} + \frac{1}{-10} = \frac{5 - 6}{60} = -\frac{1}{60} \Rightarrow v = -60\text{ cm}$$ The image is formed at $60\text{ cm}$ on the same side as the object.
Magnification: $m = \frac{v}{u} = \frac{-60}{-10} = +6$.
Nature: Virtual, erect, and highly magnified ($6$ times larger).

Ans: Given: $u = -20\text{ cm}$, $f = -20\text{ cm}$ (concave lens).
Using lens formula: $\frac{1}{v} = \frac{1}{f} + \frac{1}{u}$ $$\frac{1}{v} = \frac{1}{-20} + \frac{1}{-20} = -\frac{2}{20} = -\frac{1}{10} \Rightarrow v = -10\text{ cm}$$ The image is formed at $10\text{ cm}$ in front of the lens.
Magnification: $m = \frac{v}{u} = \frac{-10}{-20} = +0.5$. The image is virtual, erect, and diminished.

Ans: Cutting a lens horizontally along the principal axis does not change the curvature of its surfaces. Therefore, the focal length ($f$) and power ($P$) of each half remain exactly unchanged. However, because the light-collecting surface area is halved, the intensity of the resulting image will be reduced.

Ans: Lens power represents a lens's ability to bend light. When lenses are combined, the second lens further bends the refracted rays coming from the first lens. The net deviation is the sum of individual deviations, making the net power the algebraic sum of the individual powers ($P = P_1 + P_2$).

Ans: The magnification formulas are: - For a mirror: $m = -\frac{v}{u}$. - For a lens: $m = \frac{v}{u}$ (positive sign).

Ans: Given: $n_w = 1.33$ and $n_g = 1.5$.
Since glass is optically denser than water ($n_g > n_w$), light traveling from water into glass moves into a denser medium and will bend towards the normal.

Ans: Twinkling of stars is caused by atmospheric refraction.
Star light passes through multiple layers of Earth's atmosphere with fluctuating temperatures and densities (changing refractive indices). As a result, the light path is continuously refracted, shifting slightly. The apparent position of the star fluctuates rapidly, and the amount of light entering our eye varies, making the star appear to twinkle.

Ans: Critical Angle: The angle of incidence in a denser medium for which the angle of refraction in the rarer medium is exactly $90^\circ$.
Total Internal Reflection (TIR): When light traveling from a denser to a rarer medium falls at an angle of incidence greater than the critical angle, it reflects completely back into the denser medium.
Conditions: (1) Light must travel from a denser to a rarer medium. (2) Angle of incidence must exceed the critical angle.

Ans: A magnifying glass uses a convex lens.
According to lens refraction rules, a convex lens only forms a virtual, erect, and magnified image when the object is placed within its focal length (between O and F). If placed beyond F, it forms a real and inverted image, which cannot be read easily.