Question

In a Young's double slit experiment with slit separation $0.1mm$, one observes a bright fringe at angle $1/40rad$ by using light of wavelength ${\lambda }_1$. When the light of wavelength ${\lambda }_2$ is used a bright fringe is seen at the same angle in the same setup. Given that ${\lambda }_1$ and ${\lambda }_2$ are in visible range $(380nmto740nm)$, their values are:

• A. $380nm,525nm$
• B. $625nm,500nm$
• C. $380nm,500nm$
• D. $400nm,500nm$

Answer 1

The correct option is B

$625nm,500nm$

Step 1: Given:

Angle, ${\theta }_n=\frac{1}{40}$

Wavelengths: ${\lambda }_1$ and ${\lambda }_2$

Slit separation, $$\begin{gathered} d=0.1mm \\ =0.1\times {10}^{-3}m \end{gathered}$$

Step 2: Find the equation for $\lambda$:

The angle at which the bright fringe is formed is given by,

$$\begin{gathered} {\theta }_n=\frac{n\lambda }{d} \\ \frac{1}{40}=\frac{n\lambda }{0.1\times {10}^{-3}} \\ n\lambda =0.25\times {10}^{-5}m \\ \lambda =\frac{0.25\times {10}^{-5}}{n} \end{gathered}$$

Step 3: Substitute various values of n

$$\begin{gathered} n=1,\lambda =2500nm \\ n=2,\lambda =1250nm \\ n=3,\lambda =833.3nm \\ n=4,\lambda =625nm \\ n=5,\lambda =500nm \\ n=6,\lambda =416.6nm \\ n=7,\lambda =357.14nm \end{gathered}$$

Therefore, option B is the correct option.