Vardaan Learning Institute

Ans: (a) Always
For any reflecting surface, $\angle i = \angle r$ always holds true.

Ans: (d) Between the pole of the mirror and its principal focus
This is the only case where a concave mirror forms a virtual, erect, and magnified image.

Ans: (d) Clay
A lens must be transparent to allow light to pass through. Clay is opaque.

Ans: (c) Dioptre
S.I unit of power is Dioptre (D).

Ans: (b) At twice the focal length (2F)
If object is at 2F, image is at 2F, real, inverted, and of the same size.

Ans: (d) Infinity
A plane mirror can be considered as a spherical mirror of infinite radius of curvature, so $f = R/2 = \infty$.

Ans: (c) +1
Image formed is virtual, erect (+) and of same size (1) as the object.

Ans: (b) A convex mirror
A convex mirror has a wider field of view and forms diminished images, allowing full view of large objects.

Ans: (c) A convex lens of focal length 5 cm
For reading, we need a simple microscope (magnifying glass) which is a convex lens of small focal length (high power) to get high magnification ($M = 1 + D/f$).

Ans: (d) Vacuum
Light travels fastest in vacuum ($3 \times 10^8$ m/s).

Ans: (a) Both A and R are true and R is the correct explanation of A.
Bulb is placed at Focus. Light rays coming from Focus became parallel after reflection.

Ans: (b) Both A and R are true but R is not the correct explanation of A.
A ray passing through optical center suffers no deviation. R defines location of Optical Center, but doesn't explicitly explain the property of no deviation (which is due to the lens surfaces being parallel at that point).

Ans: (b) Both A and R are true but R is not the correct explanation of A.
Both statements are true. R defines refractive index, but doesn't explain why diamond specifically has 2.42 (which depends on optical density).

Ans: (a) Both A and R are true and R is the correct explanation of A.
Bending of light (refraction) causes the apparent displacement of the pencil.

Given $n_g = 1.50$, $c = 3 \times 10^8$ m/s.
Formula: $n_g = \frac{c}{v}$
$v = \frac{c}{n_g} = \frac{3 \times 10^8}{1.50} = 2 \times 10^8$ m/s.

$u = -20$ cm, $R = +30$ cm, $f = +15$ cm, $h_o = 5$ cm.
$\frac{1}{v} + \frac{1}{u} = \frac{1}{f} \Rightarrow \frac{1}{v} - \frac{1}{20} = \frac{1}{15}$
$\frac{1}{v} = \frac{1}{15} + \frac{1}{20} = \frac{7}{60}$
$v = \frac{60}{7} = 8.6$ cm (Behind mirror, Virtual).
$m = -\frac{v}{u} = +0.43$ (Diminished, Erect).
$h_i = 0.43 \times 5 = 2.15$ cm.

Object between C and F forms Image beyond C.
Nature: Real, Inverted, Magnified.
(Diagram required showing rays from object converging beyond C).

1. The incident ray, the refracted ray and the normal all lie in the same plane.
2. Snell's Law: $\frac{\sin i}{\sin r} = \text{constant} (n)$.

Light rays from the bottom of the pool travel from water (denser) to air (rarer). They bend away from the normal. When these diverging rays reach the eye, they appear to come from a point above the actual bottom, making it appear raised (shallower).

(i) Glass Slab: If the ray of light falls normally (perpendicularly) on the surface ($i=0, r=0$).
(ii) Lens: If the ray of light passes through the optical center of the lens.

(i) Convex Lens: $u = -10$ cm, $f = +15$ cm (Object within focal length).
$\frac{1}{v} - \frac{1}{u} = \frac{1}{f} \Rightarrow v = -30$ cm. Virtual, Erect, Magnified.

(ii) Concave Lens: $f = -15$ cm, $v = -10$ cm.
$\frac{1}{v} - \frac{1}{u} = \frac{1}{f} \Rightarrow u = -30$ cm.
$m = \frac{v}{u} = \frac{-10}{-30} = +0.33$ (Diminished, Erect).

(i) Power $P = \frac{1}{f(\text{in meters})}$. Degree of convergence/divergence.
(ii) $P_1 = +3.5$ D, $P_2 = -2.5$ D.
Total Power $P = +1.0$ D.
Focal length $f = \frac{1}{P} = 1$ m.

(i) Erect image and wider field of view.
(ii) Convex mirror.
(iii) $f = R/2 = 10$ cm.

(i) At Infinity.
(ii) Convex Lens.
(iii) $f = 25$ cm $= 0.25$ m. $P = 1/f = 1/0.25 = +4$ D.