JEE MAIN - Physics (2007 - No. 7)
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
$${3 \over 4}$$
$${1 \over {\sqrt 2 }}$$
$${{\sqrt 3 } \over 2}$$
$${1 \over 2}$$
Explanation
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
$$I = {I_0}{\cos ^2}\left( {{\phi \over 2}} \right)\,\,$$ where $${I_0}$$ is the maximum intensity.
NOTE : This formula is applicable when $${I_1} = {I_2}.$$
Here $$\phi = {\raise0.5ex\hbox{$\scriptstyle \pi $} \kern-0.1em/\kern-0.15em \lower0.25ex\hbox{$\scriptstyle 3$}}$$
$$\therefore$$ $${I \over {{I_0}}} = {\cos ^2}{\pi \over 6} = {\left( {{{\sqrt 3 } \over 2}} \right)^2} = {3 \over 4}$$
$$I = {I_0}{\cos ^2}\left( {{\phi \over 2}} \right)\,\,$$ where $${I_0}$$ is the maximum intensity.
NOTE : This formula is applicable when $${I_1} = {I_2}.$$
Here $$\phi = {\raise0.5ex\hbox{$\scriptstyle \pi $} \kern-0.1em/\kern-0.15em \lower0.25ex\hbox{$\scriptstyle 3$}}$$
$$\therefore$$ $${I \over {{I_0}}} = {\cos ^2}{\pi \over 6} = {\left( {{{\sqrt 3 } \over 2}} \right)^2} = {3 \over 4}$$
Comments (0)
