Question

The maximum intensity in Young's double-slit experiment is $I_0$. Distance between the slits is $d=5\lambda$, where $\lambda$ is the wavelength of monochromatic light used in the experiment. What will be the intensity of light in front of one of the slits on a screen at a distance $D=10d$?

• A. $\frac{I_0}{2}$
• B. $\frac{3}{4}{I}_0$
• C. $I_0$
• D. $\frac{I_0}{4}$

Answer 1

Answer: (A)

$\frac{I_0}{2}$

Data: $x=\frac{d}{2}=\frac{5\lambda }{2},\beta =\frac{\lambda D}{d}=\frac{\lambda 10d}{d}=10\lambda$

Intensity in front of one slit $=I_o{\cos }^2\left(\frac{\pi x}{\beta }\right)=I_o{\cos }^2\left(\frac{\pi }{4}\right)=\frac{I_o}{2}$