Question

An object $2.4\,{\text{m}}$ in front of a lens forms a sharp image on a film $12\,{\text{cm}}$ behind the lens. A glass plate $1\,{\text{cm}}$ thick of refractive index $1.50$ is interposed between lens and the film with its plane faces parallel to film. At what distance (from lens) should the object be shifted to be in sharp focus on film.
A. $7.2\,{\text{m}}$
B. $2.4\,{\text{m}}$
C. $3.2\,{\text{m}}$
D. $5.6\,{\text{m}}$

Answer 1

First of all, we will find out the focal length of the lens and then will find the shift in the image. Then again by lens formula, we will find the new object distance. We will substitute the values and manipulate accordingly.

Complete step by step answer:
In the given problem, there are two conditions mentioned. First condition is the direct formation of an image by a lens. However, the second condition is the formation of an image by a lens after a glass is interposed between the lens and the film.

Object distance is $2.4\,{\text{m}}$ .
u = - 2.4\,{\text{m}} \\\ u = - 240\,{\text{cm}} \\\
Image distance is $12\,{\text{cm}}$ .
$v = 12\,{\text{cm}}$

Let the focal length of the lens be $f$ .
So, we use the lens formula, which is given by:
$\dfrac{1}{v} - \dfrac{1}{u} = \dfrac{1}{f}$ …… (1)
Where,
$v$ indicates image distance.
$u$ indicates object distance.
$f$ indicates focal length.

Now, substituting the required values in the equation (1), we get:
\dfrac{1}{{12}} - \dfrac{1}{{\left( { - 240} \right)}} = \dfrac{1}{f} \\\ \implies \dfrac{1}{{12}} + \dfrac{1}{{240}} = \dfrac{1}{f} \\\ \implies \dfrac{7}{{80}} = \dfrac{1}{f} \\\ f = \dfrac{{80}}{7}\,{\text{cm}} \\\
The focal length of the lens is found to be $\dfrac{{80}}{7}\,{\text{cm}}$ .
There is a condition that a glass plate is interposed between the lens and the film. So, the shift of the image is given by:
${\text{shift}} = t\left( {1 - \dfrac{1}{\mu }} \right)$ …… (2)
Where,
$t$ indicates the thickness of the plate.
$\mu$ indicates a refractive index.

Substituting the required values in the equation (2), we get:
{\text{shift}} = 1\left( {1 - \dfrac{1}{{1.5}}} \right) \\\ \implies {\text{shift}} = 1\left( {1 - \dfrac{1}{{\dfrac{3}{2}}}} \right) \\\ \implies {\text{shift}} = 1\left( {1 - \dfrac{2}{3}} \right) \\\ \implies {\text{shift}} = \dfrac{1}{3} \\\

Hence, the total shift in the image is given by:
$v' = v - \dfrac{1}{3}$ …… (3)
Where,
$v'$ indicates the new position of the image.
$v$ indicates the previous position of the image.

Now, substituting the values in equation (3), we get:
v' = v - \dfrac{1}{3} \\\ \implies v' = 12 - \dfrac{1}{3} \\\ \implies v' = \dfrac{{36 - 1}}{3} \\\ \implies v' = \dfrac{{35}}{3}\,{\text{cm}} \\\
Let the new position of the object be $u'$ .
So, we can find the new object distance using the equation (1):
Substituting the required values in that equation, we get:
$\dfrac{1}{{v'}} - \dfrac{1}{{u'}} = \dfrac{1}{f} \\\ \implies \dfrac{1}{{\left( {\dfrac{{35}}{3}} \right)}} - \dfrac{1}{{u'}} = \dfrac{1}{{\dfrac{{80}}{7}}} \\\ \implies \dfrac{3}{{35}} - \dfrac{7}{{80}} = \dfrac{1}{{u'}} \\\ \implies \dfrac{1}{{u'}} = - \dfrac{1}{{560}} \\\$
Simplifying further we get:
u' = - 560\,{\text{cm}} \\\ \therefore u' = - 5.6\,{\text{m}} \\\
Hence, the object should be shifted by a distance of $5.6\,{\text{m}}$ from the lens.

So, the correct answer is “Option D”.

Note:
While solving this problem, keep in mind that the refractive index of the glass plate will affect the formation of image after it is being interposed between the lens and the film. The shift due to interposition should be considered, without which the result obtained will be irrelevant. It is important to note that the focal of the lens will be the same throughout the numerical.