## A small object stuck on the surface of a glass sphere (n = 1.5) is viewed from the diametrically opposite position….

Q: A small object stuck on the surface of a glass sphere (n = 1.5) is viewed from the diametrically opposite position. Find transverse magnification.

Click to See Solution :
Ans: m = 3
Sol: $\displaystyle \frac{n_2}{v} – \frac{n_1}{u} = \frac{n_2 – n_1}{R}$

$\displaystyle \frac{1}{v} – \frac{1.5}{-2R} = \frac{1 – 1.5}{-R}$

v = -4 R

$\displaystyle m = \frac{n_1 v}{n_2 u }$

$\displaystyle m = \frac{1.5 (-4R)}{1 (-2R) }$

m = 3

## A parallel beam of light is incident on a lens of focal length 10 cm. A parallel slab of refractive index 1.5 ….

Q: A parallel beam of light is incident on a lens of focal length 10 cm. A parallel slab of refractive index 1.5 and thickness 3 cm is placed on the other side of the lens. Find the distance of the final image from the lens. Click to See Solution :
Sol: As rays are parallel to the principal axis, image is created by lens at the focus. By placing of glass-slab,
Shift $\displaystyle = (1-\frac{1}{\mu})t$

$\displaystyle = (1-\frac{1}{1.5}) 3= 1 cm$

Irrespective of separation,
Image is shifted to the right by 1 cm. Total distance from lens 10 + 1 = 11 cm

## A composite slab consisting of different media is placed infront of a concave mirror of radius of curvature 150 cm….

Q: A composite slab consisting of different media is placed infront of a concave mirror of radius of curvature 150 cm. The whole arrangement is placed in water.An object O is placed at a distance 20 cm from the slab. The R.I. of different media are given in the diagram. Find the position of the final image formed by the Click to See Solution :
Ans: (C)

Sol: Apparent shift in the object O dut to three slabes S1 ,S2 and S3 with respect to the medium of  μ = 4/3
is given by

Shif $\displaystyle = 45(1- \frac{1}{\frac{3/2}{4/3}}) + 24(1- \frac{1}{\frac{1}{4/3}}) + 54 (1- \frac{1}{\frac{3/2}{4/3}})$

Shif $\displaystyle = 45(1- \frac{8}{9}) + 24(1- \frac{4}{3}) + 54(1- \frac{8}{9})$

Shift = 3 cm

unet = 150 cm and ROC = 150 cm. Hence image will be formed on the object itself.

## A block with a concave mirror of radius of curvature 1 m attached to one of its sides floats …

Q: COMPREHENSION
A block with a concave mirror of radius of curvature 1 m attached to one of its sides floats, with exactly half of its length immersed in water and the other half exposed to air. Any ray originating from an object O1 and O2 (as shown in figure) floating on the surface of water first gets reflected by the mirror. This then gets refracted by the water surface if the image formed by reflection is real (I1). If the image is virtual (I2), then the reflected ray never encounters the air-water interface and hence there is no refraction. The image for the next three questions refers to the final image, formed after both the reflection and refraction (if it occurs at all) has taken place.

1: The final image formed is unique (i.e. only one image is formed) if the point object floats on the surface of
water at a distance x in front of the mirror where
(A) x is less than 50 cm
(B) x is less than 1m
(C) x is between 50 cm and 1m
(D) for any value of x

2: There is no refraction of light rays reflected from the mirror if the point object floats on the surface of water at
a distance x in front of the mirror where
(A) x is less than 50 cm
(B) x is less than 1 m
(C) x is between 50 cm and 1 m
(D) for any value of x

3: The final image is not unique (i.e. more than one image for the same object) if the point object is placed a distance ëyí above the surface of water and a distance x in front of the mirror (rays are not paraxial), where y = 20 cm and x is :
(A) less than 50 cm
(B) less than 1 m
(C) between 50 cm and 1 m
(D) for any value of

Click to See Solution :
Ans: 1: (D) ; 2:(A) ; 3: (D)
Sol: 1: There is onlyone point image corresponding to a point object, as long as the object lies on the water surface
(principal axis of the mirror).Any object lying at some distance from the princpal axis results in multiple
image points.

2. If light rays diverge outward (forming a virtual image behind the mirror) after reflection, there is no refraction
at water surface after reflection. This is the case when the object lies between the focus and the