Question

In Young's double-slit experiment , the intensity of light at a point on the screen where the path difference is $\lambda$ is K units. What is the intensity of light at a point where the path difference is $\lambda /3;$ $\lambda$ being the wavelength of light used?

• A. K/4
• B. K/3
• C. K/2
• D. K

Answer 1

The correct option is A

K/4

Path difference $△=\lambda .$ Therefore, phase difference $\varphi =\frac{2\pi }{\lambda }△=2\pi .$ Hence intensity at a point where $△=\lambda or\varphi 2\pi$ is
$=I_1+I_2+2\sqrt{I_1{I}_2}cos\varphi =I_1+I_2+2\sqrt{I_1{I}_2}cos2\pi ==I_1+I_2+2\sqrt{I_1{I}_2}=I+I+2I=4I=Kunits(\because I_1=I_2=I)i.e.I=K/4.$ The intensity at a point where the path difference is
${△}^{{}^{\prime }}=\frac{\lambda }{3}and{\varphi }^{{}^{\prime }}=\frac{2\pi }{\lambda }{△}^{{}^{\prime }}=\frac{2\pi }{\lambda }\times \frac{\lambda }{3}=\frac{2\pi }{3}is{I}^{\prime }=I+I+2Icos\frac{2\pi }{3}=2I-I=I=\frac{K}{4}units$