Courses
Courses for Kids
Free study material
Offline Centres
More
Store Icon
Store
seo-qna
SearchIcon
banner

A lens forms a sharp image of a real object on the screen. On inserting a parallel slide between the lens and the screen with its thickness along the principal axis of the lens, it is found necessary to shift the screen parallel to itself distance \[d\] away from the lens for getting the image sharply focussed on it. If the refractive index of the glass relative to air is \[\mu \], the thickness of the slab is:
A) $\dfrac{d}{\mu }$
B)$\mu d$
C) $\dfrac{{\mu d}}{{\mu - 1}}$
D) $(\mu - 1)\dfrac{d}{\mu }$

Answer
VerifiedVerified
232.2k+ views
Hint: The image formed after refraction of the lens will be on the screen. To move the screen is to say that the image is formed at a different point. We will use the formula of lateral shift due to the slide to find the thickness of the slab.
Formula used: In this solution, we will use the following formula:
-Lateral shift due to a slide: $d = t\left( {1 - \dfrac{1}{\mu }} \right)$ where $t$ is the thickness of the slide and $\mu $ is the refractive index of the slide.

Complete step by step answer:
We’ve been given that when we put a slide between the lens and the screen with its thickness along the principal axis of the lens, the screen where the image is formed has to be shifted to form a sharp image.
This is equivalent to saying the position of the image is shifted. The lateral shift when a slide of thickness $t$ is placed in front of a lens is calculated as
$d = t\left( {1 - \dfrac{1}{\mu }} \right)$
To calculate $t$, we can rearrange the above equation and write
$t = \dfrac{d}{{1 - \dfrac{1}{\mu }}}$
We can further simplify it and write
$t = \dfrac{{d\mu }}{{\mu - 1}}$
Hence the thickness of the slab when the lateral shift is $d$ will be $d = t\left( {1 - \dfrac{1}{\mu }} \right)$.

So the correct choice will be option (C).

Note: The length of the slab will be the dimension of the slab that is parallel to the principal axis of the optical system. In this case, it is the length $d$. The glass slab will shift the image away than it is formed without the slab however we are only concerned with the magnitude of the shift in this question.