Question

An object 2.4 m in front of a lens forms a sharp image on a film $12cm$ behind the lens. A glass plate $1cm$ thick, of refractive index $1.5$ has been interposed between lens and films with its plane faces parallel to film. What is the distance from the lens that the object should be shifted in order to have a sharp focus on film?
$\begin{align}
  & A.7.2m \\
 & B.2.4m \\
 & C.3.2m \\
 & D.5.6m \\
\end{align}$

Answer 1

Hint: In order to solve this question formula of lens should be used. Here the distance of the object and the image from the lens is mentioned in the question. Using the lens formula, arrive at the focus of length and then find the normal shift which has taken place for the image.

Complete step by step answer:

For this question it has given that,
$\begin{align}
  & u=-240cm \\
 & v=12cm \\
 & n=1.5 \\
 & t=1cm \\
\end{align}$
In which \[u\] is the object distance from the lens,\[v\] is the image distance,\[n\] is the refractive index of the glass plate and \[t\] is the thickness of the glass plate.
So first of all we have to find the focal length of the glass plate. For that,
Using the lens formula,
\[\dfrac{1}{f}=\dfrac{1}{v}-\dfrac{1}{u}\]
Substituting the terms in this equation will give,
\[\begin{align}
  & \dfrac{1}{f}=\dfrac{1}{12}-\dfrac{1}{-240} \\
 & \dfrac{1}{f}=\dfrac{1}{12}+\dfrac{1}{240} \\
\end{align}\]
Simplifying this can be written as,
\[f=\dfrac{80}{7}cm\]
The normal shift in the glass slab can be written as,
$s=t\left( 1-\dfrac{1}{n} \right)$
Substituting the terms in this equation will give,
$s=1\left( 1-\dfrac{1}{1.5} \right)=\dfrac{1}{3}$
This should be subtracted from the image distance,\[v\],
Therefore we can write that,
\[v=12-\dfrac{1}{3}=\dfrac{35}{3}\]
Using this we can find the object distance,
Using lens formula once again will give,
$\begin{align}
  & \dfrac{1}{u}=\dfrac{1}{v}+\dfrac{1}{f} \\
 & \dfrac{1}{u}=\dfrac{3}{35}+\dfrac{7}{80} \\
 & u=5.6m \\
\end{align}$

So, the correct answer is “Option D”.

Note: The Apparent shift of the ray is given as the perpendicular distance between the incident and the emergent ray which is related to the refraction of light through a glass slab. When refraction is taking place, the apparent shift happening in the position of the object along the direction of normal is called normal shift.