Q170 / 170 · Manipal (MET) 2015 · Physics
In Young's double slit experiment, fringes of width $\beta$ are produced on the screen kept at a distance of 1 m from the slit. When the screen is moved away by $5 \times 10^{-2} \mathrm{~m}$, fringe width changes by $3 \times 10^{-5} \mathrm{~m}$. The separation between the slits is $1 \times 10^{-3} \mathrm{~m}$. The wavelength of light used is......nm.
- A$500$
- Correct answerB$600$
- C$700$
- D$400$
Solution
Since, we know that in Young's double slit experiment.
$\beta=\frac{D \lambda}{d}$ ...(i)
$\beta^{\prime}=\frac{D^{\prime} \lambda}{d}$ ...(ii)
Subtract Eq. (ii) from Eq. (i), we get
$\beta-\beta^{\prime}=\frac{\lambda\left(D-D^{\prime}\right)}{d}$
Substituting the given values; we get
$3 \times 10^{-5}=\frac{\lambda \times 5 \times 10^{-2}}{1 \times 10^{-3}}$
$\begin{array}{ll}\Rightarrow & \lambda=\frac{3 \times 10^{-8}}{5 \times 10^{-2}}=0.6 \times 10^{-6} \mathrm{~m} \\ \Rightarrow & \lambda=600 \times 10^{-9} \mathrm{~m} \\ \Rightarrow & \lambda=600 \mathrm{~nm}\end{array}$
Hence, the correct answer is option: 'B'.