
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
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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.
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.
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