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A thin glass plate of thickness is $\dfrac{{2500}}{3}\lambda $ ( $\lambda $ is wavelength of light used) and refractive index $\mu = 1.5$ is inserted between one of the slits and the screen in Young’s double slit experiment. At a point on the screen equidistant from the slits, the ratio of the intensities before and after the introduction of the glass plate is:
A) $2:1$
B) $1:4$
C) $4:1$
D) $4:3$

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Last updated date: 25th Feb 2024
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Hint: Calculate the path difference due to insertion of the thin glass plate. Due to this path difference, there will be a phase difference. Calculate that path difference. Now, intensity can be calculated using the formula of resultant intensities.

Complete step by step solution:
We know that there will be path difference due to the insertion of the glass plate and this path difference will also lead to phase difference.
Now, the path difference when the thin glass plate is inserted is given as $\Delta p$ ,
$\Delta p = (\mu - 1)t$.....................equation 1
Here $t$ is the thickness of the glass plate
$\mu $ is the refractive index of the glass
Now, the phase difference when the thin glass plate is inserted $$\Delta \phi $$ is given as:
$$\Delta \phi = \dfrac{{2\pi }}{\lambda }\Delta p$$
$$\Delta \phi = \dfrac{{2\pi }}{\lambda }\left( {\mu - 1} \right)t$$.............equation 2
Where $\lambda $ is wavelength of light used
The intensity at the centre $${I_c}$$ after the insertion of the glass plate will be given as:
$${I_c} = {I_s} + {I_s} + 2\sqrt {{I_s}^2} \cos \Delta \phi $$
Here, $${I_s}$$ is the intensity of light from each slit
Solving this equation, we get
$${I_c} = 2{I_s}(1 + \cos \Delta \phi )$$
But from trigonometric identities, we know that $$1 + \cos \theta = 2{\cos ^2}\left( {\dfrac{\theta }{2}} \right)$$
$$ \Rightarrow {I_c} = 2{I_s}(2{\cos ^2}\left( {\dfrac{{\Delta \phi }}{2}} \right))$$
$$ \Rightarrow {I_c} = 4{I_s}({\cos ^2}\left( {\dfrac{{\Delta \phi }}{2}} \right))$$
Before the glass plate was inserted, the phase difference was zero and hence the intensity was
$${I_c} = 4{I_s}({\cos ^2}{0^0})$$
$$ \Rightarrow {I_c} = 4{I_s} = {I_0}$$............equation 3
Substituting the value of phase difference, we get
$$ \Rightarrow {I_c} = 4{I_s}{\cos ^2}\left( {\dfrac{{2\pi }}{\lambda }\left( {\mu - 1} \right)t} \right)$$
$$ \Rightarrow {I_c} = 4{I_s}{\cos ^2}\left( {\dfrac{{2\pi }}{\lambda }\left( {1.5 - 1} \right)\dfrac{{2500}}{3}\lambda } \right)$$
$$ \Rightarrow {I_c} = 4{I_s}{\cos ^2}\left( {2\pi \left( {\dfrac{1}{2}} \right)\dfrac{{2500}}{3}} \right)$$
$$ \Rightarrow {I_c} = 4{I_s}{\cos ^2}\left( {2500\dfrac{\pi }{3}} \right)$$
But
$${\cos ^2}\left( {2500\dfrac{\pi }{3}} \right) = \dfrac{1}{4}$$
$$ \Rightarrow {I_c} = 4{I_s} \times \dfrac{1}{4}$$
$$ \Rightarrow {I_c} = {I_s}$$
Comparing this with equation $3$ , we have
$$\dfrac{{{I_0}}}{{{I_C}}} = \dfrac{4}{1}$$
Therefore, the ratio of the intensities before and after the introduction of the glass plate is $$4:1$$

Thus, option C is the correct option.

Note: In Young’s double slit experiment, a pattern of bright and dark fringes is observed. Remember when there is no glass plate, the phase difference and the path difference will be zero. Also remember that due to insertion of a thin glass plate, the intensity decreased for this problem.