Question

A beam of light is incident at point $I$ on a screen, a plane glass plate of thickness $t$ and refractive index $n$ is placed in the path of light. The displacement of point will be:
(A) $t\left( {1 - \dfrac{1}{n}} \right)$ times nearer
(B) $t\left( {1 + \dfrac{1}{n}} \right)$ times nearer
(C) $t\left( {1 - \dfrac{1}{n}} \right)$ times farther
(D) $t\left( {1 + \dfrac{1}{n}} \right)$ times farther

Answer 1

The speed of light inside a medium is less than the speed of light in the air. So the light will travel less distance within the glass, and the difference in the distance covered by the light within the glass is equal to the required shift.

Formula used: The formula used to solve this question is given by
$\dfrac{{h'}}{h} = \mu$ , here $h'$ is the apparent height of an object as viewed by an observer inside a medium of refractive index $\mu$ , and the real height of the object is equal to $h$ .

Complete step-by-step solution
According to the question, the given beam of light is converging at a point $1$ on the screen. But then a plane glass of refractive index $n$ and of thickness $t$ is introduced in between the path of the light.
So in the first case, the beam was travelling in the air, but in the second case the beam of light first travels in air and then passes through the glass and after travelling through the glass, it converges at a point in the air. Let us call that point as $I'$ . We know that the speed of light in the glass is less than the speed of light in the air. So the speed of the beam of light will be lowered within the thickness of the glass plate. Because of this, the beam of light will converge at a point located farther away from the point $1$ .
So the options A and B are incorrect.
Now, we know that the apparent depth of a point inside a transparent medium is given by
$\mu = \dfrac{d}{{d'}}$
Substituting $d = t$ in the above equation, we get
$\mu = \dfrac{t}{{d'}}$
$d' = \dfrac{t}{\mu }$
Since the speed of light is less in the glass, it travels this much distance within the glass. But in the air, it covers the distance equal to the thickness of the plate, that is, $t$ . So the shift is given by
$x = t - \dfrac{t}{\mu }$
$\Rightarrow x = t\left( {1 - \dfrac{1}{n}} \right)$
Thus, the point $I'$ is $t\left( {1 - \dfrac{1}{n}} \right)$ times farther away from the point $I$ .
Hence, the correct answer is option C.

Note:
We could also use the lens equation to solve this question. For this we need to apply the lens equation two times. The object will be virtual in this case as the beam of light incident on the lens is converging.