In a Young’s double slit experiment the intensity at a point where the path difference is $${\frac{\lambda }{6}}$$ ($$\lambda $$ being the wavelength of light used) is $$I.$$ If $${{I_0}}$$ denotes the maximum intensity, $$\frac{I}{{{I_0}}}$$ is equal to
A.
$$\frac{3}{4}$$
B.
$$\frac{1}{{\sqrt 2 }}$$
C.
$$\frac{{\sqrt 3 }}{2}$$
D.
$$\frac{1}{2}$$
Answer :
$$\frac{3}{4}$$
Solution :
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
$$I = {I_0}{\cos ^2}\left( {\frac{\phi }{2}} \right)$$
where $${I_0}$$ is the maximum intensity. NOTE : This formula is applicable when $${I_1} = {I_2}.$$ Here $$\phi = \frac{\pi }{3}$$
$$\eqalign{
& \therefore \,\,\frac{I}{{{I_0}}} = {\cos ^2}\frac{\pi }{6} = {\left( {\frac{{\sqrt 3 }}{2}} \right)^2} \cr
& = \frac{3}{4} \cr} $$
Releted MCQ Question on Optics and Wave >> Wave Optics
Releted Question 1
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