Question 425: Waves Optics

Question

In the Young's double slit experiment, the intensity of light at a point on the screen where the path difference $$\lambda $$ is K, ($$\lambda $$ being the wavelength of light used). The intensity at a point where the path difference is $$\lambda $$/4 will be

Accepted Answer

Phase difference, $$\phi $$ = $${{2\pi } \over \lambda }$$ $$ 	imes $$ Path difference
When path difference is $$\lambda $$, then
$$\phi $$ = $${{2\pi } \over \lambda } 	imes \lambda $$ = 2$$\pi $$
$$ 	herefore $$ I = 4I
0
$${\cos ^2}\left( {{{2\pi } \over 2}} ight)$$ = 4I
0
cos
2
($$\pi $$) = 4I
0
= K ....(1)
When path difference is , $${\lambda  \over 4}$$
 then
$$\phi $$ = $${{2\pi } \over \lambda } 	imes {\lambda  \over 4}$$ = $${{\pi  \over 2}}$$
$$ 	herefore $$ I = 4I
0
$${\cos ^2}\left( {{{\pi } \over 4}} ight)$$ = 2I
0
= $${K \over 2}$$

Answer

K

Answer

K/4

Answer

K/2

Answer

zero

Waves Optics

In the Young's double slit experiment, the intensity of light at a point on the screen where the path difference $\lambda is K, (\lambda being the wavelength of light used). The intensity at a point where the path difference is \lambda $/4 will be

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