Question 13: Waves Optics

Question

Young's double slit experiment is performed in a medium of refractive index of 1.33 . The maximum intensity is $ I_{0} $. The intensity at a point on the screen, where path difference between the light coming out from slits is $ \frac{\lambda}{4} $, is

Accepted Answer

In Young's double slit experiment within a medium with a refractive index of 1.33, the maximum intensity is represented as $ I_0 $.
To find the intensity at a point on the screen where the path difference between the light from the slits is $ \frac{\lambda}{4} $, follow these steps:
Path Difference
: Given as $ \Delta x = \frac{\lambda}{4} $.
Phase Difference Calculation
:
$    \phi = \left(\frac{2\pi}{\lambda}ight) 	imes \Delta x    $
Substituting the given path difference:
$    \phi = \frac{2\pi}{\lambda} 	imes \frac{\lambda}{4} = \frac{\pi}{2}    $
Intensity Calculation
: The intensity at any point on the screen with a phase difference $ \phi $ is given by:
$    I = I_0 \cos^2\left(\frac{\phi}{2}ight)    $
Substituting the phase difference:
$    I = I_0 \cos^2\left(\frac{\pi/2}{2}ight)    $
Simplifying further:
$    I = I_0 \cos^2\left(\frac{\pi}{4}ight)    $
$    I = I_0 \left(\frac{1}{\sqrt{2}}ight)^2    $
$    I = \frac{I_0}{2}    $
Thus, the intensity at the specified point on the screen is $\frac{I_0}{2}$.

Answer

0

Answer

$ \frac{I_{0}}{2} $

Answer

$ \frac{3 I_{0}}{8} $

Answer

$ \frac{2 I_{0}}{3} $

Waves Optics

Young's double slit experiment is performed in a medium of refractive index of 1.33 . The maximum intensity is I_{0} . The intensity at a point on the screen, where path difference between the light coming out from slits is \frac{\lambda}{4} , is

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