Question

White coherent light $(4000-7000\stackrel{\circ }{A})$ is sent through the slits of a YDSE. The separation between the slits is $0.5\text{mm}$ and screen is $50\text{cm}$ away from the slits. There is a hole in the screen at a point 1.0 mm away (along the width of the fringe) from the central line.
a. Which wavelength(s) will be absent in the light coming from the hole?
b. Which wavelength(s) will have a strong intensity?

• A. $5000\stackrel{\circ }{A}$, $7000\stackrel{\circ }{A}$
• B. $4000\stackrel{\circ }{A}$, $6000\stackrel{\circ }{A}$.
• C. $7000\stackrel{\circ }{A}$, $5000\stackrel{\circ }{A}$.
• D. $3000\stackrel{\circ }{A}$, $6000\stackrel{\circ }{A}$.

Answer 1

The correct option is B $4000\stackrel{\circ }{A}$, $6000\stackrel{\circ }{A}$.

Given, White coherent light range=$(4000\stackrel{\circ }{A}-7000\stackrel{\circ }{A})$


Total path difference $\Delta x=\frac{yd}{D}$
from the given information,
$\Delta x=\frac{1mm\times 5mm}{10\times 500mm}=\frac{1}{1000}mm$

For the dark fringe =$\frac{2n+1}{2}\lambda$
for $\lambda =4000\stackrel{\circ }{A}$
so, $\frac{1}{1000}mm=\frac{2n+1}{2}\lambda \frac{{10}^7}{{10}^3}=\frac{2n+1}{2}\times 4000\Rightarrow n=2$

for $\lambda =7000\stackrel{\circ }{A}$
$\frac{{10}^7}{{10}^3}=\frac{2n+1}{2}\times 7000\Rightarrow n=\frac{13}{14}$
where, $n<1$
hence, n should be 1 or less then 1.

For, $n=1$
$\frac{{10}^7}{{10}^3}=\frac{2+1}{2}\lambda \lambda =\frac{20000}{3}=6666.66\stackrel{\circ }{A}$

Hence, the limits for wevlenght $\lambda$ are between $4000\stackrel{\circ }{A}$ and $6666.66\stackrel{\circ }{A}$.