Question

A parallel beam of light $(\lambda =5000\stackrel{\circ }{A})$ is incident at an angle $\alpha ={30}^{\circ }$ with the normal to the slit plane in a Young’s double slit experiment. Assume that the intensity due to each slit at any point on the screen is $I_0$. Point O is equidistant from $s_1$ and $s_2$. The distance between slits is 1mm and between slits and screen is 3m.

• A. The intensity at O is $4I_0$
• B. The intensity at O is zero.
• C. The intensity at a point on the screen 4m from O is $4I_0$
• D. The intensity at a point on the screen 4m from O is zero.

Answer 1

Answer: (A), (C)

The correct options are
A The intensity at O is $4I_0$
C The intensity at a point on the screen 4m from O is $4I_0$


For O
$\Delta x=dsin{\theta }_0$
$\Delta x=1\left(\frac{1}{2}\right)$
$\Delta x=0.5mm$ and $\lambda =5000\stackrel{\circ }{A}$
$\therefore \Delta x=n\lambda$ (with value of n = integer)
So, at O maxima occurs,
and $I=4I_0$
at $Y=4m$
$\Delta x=d(sin\theta -sin{\theta }_0)or\Delta x=d(sin\theta +sin{\theta }_0)$
$=1(\frac{4}{5}-\frac{1}{2})or1(\frac{4}{5}+\frac{1}{2})$
$\Delta x=0.3mm$ or $\Delta x=1.3mm$
Again these are multiple of $\lambda$
So, maxima at these points.