Question 168: Wave Optics

Question

In a Young's double slit experiment the intensity at a point where the path difference is $${\lambda  \over 6}$$ ( $$\lambda $$ being the wavelength of light used ) is $$I$$. If $${I_0}$$ denotes the maximum intensity, $${I \over {{I_0}}}$$ is equal to

Accepted Answer

The intensity of light at any point of the screen  where the phase difference due to light coming from the two slits is $$\phi $$ is  given by
$$I = {I_0}{\cos ^2}\left( {{\phi  \over 2}} ight)\,\,$$ where $${I_0}$$ is the maximum intensity.
NOTE :
This formula is applicable when $${I_1} = {I_2}.$$
Here $$\phi  = {aise0.5ex\hbox{$\scriptstyle \pi $}
\kern-0.1em/\kern-0.15em
\lower0.25ex\hbox{$\scriptstyle 3$}}$$
$$	herefore$$ $${I \over {{I_0}}} = {\cos ^2}{\pi  \over 6} = {\left( {{{\sqrt 3 } \over 2}} ight)^2} = {3 \over 4}$$

Answer

$${3 \over 4}$$

Answer

$${1 \over {\sqrt 2 }}$$

Answer

$${{\sqrt 3 } \over 2}$$

Answer

$${1 \over 2}$$

Wave Optics

In a Young's double slit experiment the intensity at a point where the path difference is ${\lambda \over 6} ( \lambda being the wavelength of light used ) is I. If {I_0} denotes the maximum intensity, {I \over {{I_0}}}$ is equal to

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