Question

In YDSE, $D=1m$, $d=1mm$, and $\lambda =500nm$. The distance of 100th maxima from the central maxima is

• A. $\frac{1}{2}m$
• B. $\frac{\sqrt{3}}{2}m$
• C. $\frac{1}{\sqrt{3}}m$
• D. $\frac{1}{20}m$

Answer 1

Answer: (D)

$\frac{1}{20}m$

Position of the $nth$ bright fringe, $y=n\lambda \frac{D}{d}$

For central maxima, $n=0⟹y_1=0$

For ${100}^{th}$ maxima, $y_2=100\times 500\times {10}^{-9}\frac{1}{{10}^{-3}}m$

$⟹y_2=5cm$

Thus distance of 100th maxima from central maxima $y=5cm=1/20m$