## A mark is made on the surface of a glass sphere of diameter 10cm and refractive index 1.5. It is viewed through…..

Q. A mark is made on the surface of a glass sphere of diameter 10cm and refractive index 1.5. It is viewed through the glass from a portion directly opposite. The distance of the image of the mark from the centre of the sphere will be

(a) 15 cm

(b) 17.5 cm

(c) 20 cm

(d) 22.5 cm

Ans: (a)

Sol: μ2/v -μ1/u = (μ21)/R

1/v -1.5/(-10) = (1-1.5)/(-5)

1/v + 3/20 = 1/10

1/v = -3/20 +1/10 = -1/20

v= -20 cm

Hence , Distance from centre = 15 cm

## The critical angle for refraction from glass to air is 30° and that from water to air is 37°. Find the critical angle…..

Q. The critical angle for refraction from glass to air is 30° and that from water to air is 37°. Find the critical angle for refraction from glass to water

(a)sin-1(5/6)

(b) 51°3′

(c) 61°2′

(d) 63°3′

Ans: A

Refraction from glass to air , critical angle = 30°

μg sin30° = μa sin90°

Refraction from water to air , critical angle = 37°

μw sin37° = μa sin90°

Let critical angle for Refraction from glass to water = θ

sinθ = sin30°/sin37° = (1/2)/(3/5) =5/6

θ = sin−1(5/6)

## The critical angle of light going from medium A into medium B is θ. The speed of light in medium A is v…..

Q. The critical angle of light going from medium A into medium B is θ. The speed of light in medium A is v. The speed of light in medium B is

(a) v/sinθ

(b) v sinθ

(c) v/tanθ

(d) v tanθ

Ans: (a)

Sol: μAsinθ = μBsin90

sinθ = μBA    …..(i)

vA/vB = (c/μB)/(c/μA) = μBA = sinθ   from(i)

vB = vA/sinθ = v/sinθ

## A small air bubble is inside a transparent cube of side length 24cm and of refractive index 4/3. If the apparent…..

Q. A small air bubble is inside a transparent cube of side length 24cm and of refractive index 4/3. If the apparent distance air bubble from a face is 9cm then its apparent distance from opposite face is

(a) 6cm

(b) 8cm

(c) 9 cm

(d) 12cm

Ans: (c)

Sol: Refractive index , μ = Real distance / Apparent distance

Real distance = μ×Apparent distance = 9×4/3 = 12 cm

Apparent distance from opposite face = Real distance / μ = 12/(4/3) = 36/4 = 9 cm

## A linear object is placed along the axis of a mirror as shown in figure. If ‘f’ is the focal length of the mirror then the length of image is.

Q. A linear object is placed along the axis of a mirror as shown in figure. If ‘f’ is the focal length of the mirror then the length of image is. (a) (2f)/3
(b) f
(c) f/3
(d) None
Ans: b

Sol: Let AB be linear object & A is nearer to pole .

For nearer end A , Applying mirror formula

1/v + 1/u = 1/f

1/v + 1/(-3/2)f = -1/f

1/v = 2/3f -1/f = -1/3f

v = -3f

For end B ,

1/v + 1/u = 1/f

1/v – 1/2f = -1/f

1/v = 1/2f – 1/f = -1/2f

v= -2f

As Length of object = 2f – 3f/2 = f/2

Length of image = 3f-2f = f