Question

In Young’s double slit experiment, one of the slits is wider than the other, so that the amplitude of light from one slit is double of that from the other slit. If \[{{I}_{m}}\] is the maximum density, what is the resultant intensity when they interfere at phase difference \[Q\]?
A. \[\dfrac{{{I}_{m}}}{9}\left( 1-8{{\cos }^{2}}\dfrac{Q}{2} \right)\]
B. \[\dfrac{{{I}_{m}}}{9}\left( 1+8{{\cos }^{2}}\dfrac{Q}{2} \right)\]
C. \[\dfrac{{{I}_{m}}}{9}\left( 1-8{{\cos }^{2}}Q \right)\]
D. \[\dfrac{{{I}_{m}}}{9}\left( 1-{{\sin }^{2}}\dfrac{Q}{2} \right)\]

Answer 1

Hint:Apply the formula for resultant intensity i.e., \[{{I}_{r}}={{I}_{1}}+{{I}_{2}}+2\sqrt{{{I}_{1}}{{I}_{2}}}\cos Q\] and thereafter apply \[{{I}_{m}}={{\left( {{a}_{1}}+{{a}_{2}} \right)}^{2}}\] for calculating maximum intensity. Finally express the resultant intensity \[\left( {{I}_{r}} \right)\] in terms of maximum intensity\[\left( {{I}_{\operatorname{m}}} \right)\].

Complete step by step answer:

Given that the amplitude of light from one slit is double of that from the other slit so,

Let \[{{a}_{1}}=a\], \[{{I}_{1}}={{a}_{1}}^{2}={{a}^{2}}\]
Now, \[{{a}_{2}}=2a\], \[{{I}_{2}}={{a}_{2}}^{2}={{\left( 2a \right)}^{2}}\]
Therefore, \[{{I}_{2}}=4{{I}_{1}}\]

Using formula for resultant intensity,

\[{{I}_{r}}={{I}_{1}}+{{I}_{2}}+2\sqrt{{{I}_{1}}{{I}_{2}}}\cos Q\]
\[\Rightarrow {{I}_{r}}={{I}_{1}}+4{{I}_{1}}+2\sqrt{4{{I}_{1}}^{2}}\cos Q\]
\[\Rightarrow {{I}_{r}}=5{{I}_{1}}+4{{I}_{1}}\cos Q\]-------- (1)
Now, \[{{I}_{m}}={{\left( {{a}_{1}}+{{a}_{2}} \right)}^{2}}\]\[={{(a+2a)}^{2}}\]\[=9{{a}^{2}}\]
Here, \[{{I}_{\operatorname{m}}}=9{{I}_{1}}\] \[\Rightarrow {{I}_{1}}=\dfrac{{{I}_{\operatorname{m}}}}{9}\]--------(2)
Putting value of \[{{I}_{1}}\] in equation (1),
\[{{I}_{r}}=\dfrac{5{{I}_{\operatorname{m}}}}{9}+\dfrac{4{{I}_{\operatorname{m}}}}{9}\cos Q\]
\[{{I}_{r}}=\dfrac{{{I}_{\operatorname{m}}}}{9}\left( 5+4\cos Q \right)\]
\[{{I}_{r}}=\dfrac{{{I}_{\operatorname{m}}}}{9}\left( 5+8{{\cos }^{2}}\dfrac{Q}{2}-4 \right)\]
\[{{I}_{r}}=\dfrac{{{I}_{\operatorname{m}}}}{9}\left( 1+8{{\cos }^{2}}\dfrac{Q}{2} \right)\]

Hence, the correct option is B, i.e., \[\dfrac{{{I}_{m}}}{9}\left( 1+8{{\cos }^{2}}\dfrac{Q}{2} \right)\]

Additional Information:

Young's experiment was based on the hypothesis that if light were wave-like in nature, then it should behave in a manner similar to ripples or waves on a pond of water.
In order to test his hypothesis, Young devised an ingenious experiment. Using sunlight diffraction through a small slit as a source of coherent illumination, he projected the light rays emanating from the slit onto another screen containing two slits placed side by side. Light passing through the slits was then allowed to fall onto a screen.
Young observed that when the slits were large, spaced far apart and close to the screen, then two overlapping patches of light formed on the screen. However, when he reduced the size of the slits and brought them closer together, the light passing through the slits and onto the screen produced distinct bands of colour separated by dark regions in a serial order.

Note: Students should try to memorize the basic formulas related to Young’s double slit experiment such as relation between amplitude and intensity, expression for resultant intensity, maximum intensity etc. so that they can do this type of questions easily and in relatively less time.