Question

While a moving picture is being screened, a boy introduces a glass slab thickness $3\,{\text{cm}}$ and refractive index 1.5 between the projector and the screen. In order to have a clear picture on the screen, the screen show to be moved through a distance of:
A. $1\,{\text{cm}}$ away
B. $1\,{\text{cm}}$ nearer
C. $2\,{\text{cm}}$ away
D. $3\,{\text{cm}}$ away

Answer 1

Use the formula for the normal shift of the image due introduction of the glass slab. This formula gives the relation between thickness of the glass slab, refractive index of the air medium and the refractive index of medium (here glass medium as the glass slab is made up of glass).

Formula used:
The normal shift in the distance of image formed due to the introduction of the glass slab is given by
$d = t\left( {1 - \dfrac{{{\mu _{air}}}}{{{\mu _{medium}}}}} \right)$ …… (1)
Here, $d$ is the shift in the distance of image formed on the screen, $t$ is the thickness of the glass slab, ${\mu _{air}}$ is the refractive index of air medium and ${\mu _{medium}}$ is the refractive index of the medium.

Complete step by step solution:
We have given that the thickness of the glass slab is $3\,{\text{cm}}$ and the refractive index of the glass slab is 1.5.
$t = 3\,{\text{cm}}$
${\mu _{glass}} = 1.5$
We know that the refractive index of the air medium is 1.
${\mu _{air}} = 1$
We can determine the shift in the image distance of the image formed on the screen when the glass slab is introduced between the projector and the screen using equation (1).
Rewrite equation (1) for the glass medium of the glass slab.
$d = t\left( {1 - \dfrac{{{\mu _{air}}}}{{{\mu _{glass}}}}} \right)$
Substitute $3\,{\text{cm}}$ for $t$, 1 for ${\mu _{air}}$ and 1.5 for ${\mu _{glass}}$ in the above equation.
$d = \left( {3\,{\text{cm}}} \right)\left( {1 - \dfrac{1}{{1.5}}} \right)$
$\Rightarrow d = \left( {3\,{\text{cm}}} \right)\left( {\dfrac{{1.5 - 1}}{{1.5}}} \right)$
$\Rightarrow d = \left( {3\,{\text{cm}}} \right)\left( {\dfrac{{0.5}}{{1.5}}} \right)$
$\Rightarrow d = \left( {3\,{\text{cm}}} \right)\left( {\dfrac{1}{3}} \right)$
$\Rightarrow d = 1\,{\text{cm}}$
From the above result, we can say that the shift in the image formed on the projector is $1\,{\text{cm}}$ away from the glass slab.
Therefore, we need to shift the screen $1\,{\text{cm}}$ away from its original position.

So, the correct answer is “Option A”.

Note:
There is no need to convert the unit of thickness of the glass slab from the CGS system of units to the SI system of units as the final answer we are getting is also in the CGS system of units. The students may get confused about how we can determine whether the projector should shift 1 cm away and not nearer. As the value of the shift obtained in positive, we concluded that the projector should be shifted away. If we obtain shift value negative, then the screen needs to move nearer.