Question 415: Wave Optics

Question

The intensity at the maximum in a Young's double slit experiment is $$I$$
0
. Distance between two slits is d = 5$$\lambda $$, where $$\lambda $$ is the wavelength of light used in the expreriment. What will be the intensity in front of one of the slits on the screen placed at a distance D = 10d ?

Accepted Answer

Path difference S
2
P Ã¢ÂÂ S
1
P = $$\sqrt {{{\left( {50\lambda } ight)}^2} + {{\left( {5\lambda } ight)}^2}} $$ - 50$$\lambda $$ = 0.25$$\lambda $$
S
2
P Ã¢ÂÂ S
1
P = $${\lambda  \over 4}$$
Phase difference,
$$\Delta $$$$\phi $$ = $${{2\pi } \over \lambda } 	imes {\lambda  \over 4}$$ = $${\pi  \over 2}$$
So, resultant intensity at the desired point 'P' is
I = I
0
cos
2
$${\phi  \over 2}$$ = I
0
cos
2
$${\pi  \over 4}$$ = $${{{I_0}} \over 2}$$

Answer

$${3 \over 4}{I_0}$$

Answer

$${{{I_0}} \over 2}$$

Answer

I
0

Answer

$${{{I_0}} \over 4}$$

Wave Optics

The intensity at the maximum in a Young's double slit experiment is $I 0 . Distance between two slits is d = 5\lambda , where \lambda $ is the wavelength of light used in the expreriment. What will be the intensity in front of one of the slits on the screen placed at a distance D = 10d ?

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