Question

A point object is placed on principal axis of a concave mirror of focal length 15 cm) at a distance u = 61 cm
a pole .a slabs of thickness t=3 cm and refractive index=1.5 is placed with two sides perpendicular to
principle axis such that its nearest face is x is placed on two pole. The final image of object is to be considered after
refraction by slabs, reflection by mirror and final retraction by slab
if x= 30 cm then distance of final image from pole
1= 19, 2=21, 3= 23, 4=24​

Answer 1

Answer:

23 par bene gi image physics

Answer 2

Given that,

Focal length = 15 cm  

Distance of the pole from the mirror = 61 cm

Refractive index = 1.5  

Thickness = 3 cm

Apparent shift in position of the object due to refraction through the slab

We need to calculate the apparent distance

Using formula of thickness

$d_{0}=t(1-\dfrac{1}{\mu})$

Put the value into the formula

$d_{0}=3(1-\dfrac{1}{1.5})$

$d_{0}= 1\ cm$

The object distance will be

$u=61-1=60\ cm$

We need to calculate the image distance

Using formula of mirror

$\dfrac{1}{f}=\dfrac{1}{v}+\dfrac{1}{u}$

Put the value into the formula

$\dfrac{1}{15}=\dfrac{1}{v}+\dfrac{1}{60}$

$\dfrac{1}{v}=\dfrac{1}{15}-\dfrac{1}{60}$

$v=20\ cm$

The image formed 20 cm in right of the mirror.

We need to calculate the final image of the pole  

Using formula of distance  

$v'=v+d_{0}$

Put the value into the formula

$v'=20+1$

$v'=21\ cm$

Hence, The final image is formed at a distance 21 cm away from the mirror.

(2) is correct option.