Question

In Young’s double slit experiment , one of the slits is wider than the other, so that the amplitude of the light from one slit is double that from the other slit. If $I_{m}$be the maximum intensity, the resultant intensity when they interfere at phase difference $\varphi$is given by :

• A. $\frac{I_{m}}{3}\left( 1 + 2\cos^{2}\frac{\varphi}{2} \right)$
• B. $\frac{I_{m}}{5}\left( 1 + 4\cos^{2}\frac{\varphi}{2} \right)$
• C. $\frac{I_{m}}{9}\left( 1 + 8\cos^{2}\frac{\varphi}{2} \right)$
• D. $\frac{I_{m}}{9}\left( 8 + \cos^{2}\frac{\varphi}{2} \right)$

Answer 1

Answer: (C)

$\frac{I_{m}}{9}\left( 1 + 8\cos^{2}\frac{\varphi}{2} \right)$

Answer 2

: Here $A_{2} = 2A_{1}$

$\because Intensity \propto (Amplitude^{2}$

$\therefore\frac{I_{2}}{I_{1}} = \left( \frac{A_{2}}{A_{1}} \right)^{2} = \left( \frac{2A_{1}}{A_{1}} \right)^{2} = 4$

$I_{2} = 4I_{1}$

Maximum intensity, $I_{m} = \left( \sqrt{I_{1}} + \sqrt{I_{2}} \right)^{2}$

$I_{m} = \left( \sqrt{I_{1}} + \sqrt{4I_{1}} \right)^{2} = \left( 3\sqrt{I_{1}} \right)^{2} = 9I_{1}$

Or $I_{1} = \frac{I_{m}}{9}$…… (i)

Resulting intensity

$I = I_{1} + I_{2} + 2\sqrt{I_{1}I_{2}}\cos\varphi$

}{= 5I_{1} + 4I_{1}\cos\varphi = I_{1} + 4I_{1} + 4I_{1}\cos\varphi }{= I_{1} + 4I_{1}(1 + \cos\varphi) }{= I_{1} + 8I_{1}\cos^{2}\frac{\varphi}{2}\left( \because 1 + \cos\varphi = 2\cos^{2}\frac{\varphi}{2} \right) }{= I_{1}\left( 1 + 8\cos^{2}\frac{\varphi}{2} \right)}$$ Putting the value of $I_{1}$from eqn. (i) we get $$I = \frac{I_{m}}{9}\left( 1 + 8\cos^{2}\frac{\varphi}{2} \right)$$