Q124 / 180 · KCET 2012 · Physics

In Young's double slit experiment, fringes of width $\beta$ are produced on a screen kept at a distance of $1 \mathrm{~m}$ 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 the light used is

  • A
    $400 \mathrm{~nm}$
  • B
    $500 \mathrm{~nm}$
  • Correct answer
    C
    $600 \mathrm{~nm}$
  • D
    $700 \mathrm{~nm}$

Solution

In YDSE, we have $$ \begin{aligned} &\text { Wavelength } \begin{aligned} \lambda &=\frac{\Delta \beta \mathrm{d}}{\Delta \mathrm{D}}=\frac{3 \times 10^{-5} \times 1 \times 10^{-3}}{5 \times 10^{-2}} \\ &=6 \times 10^{-7} \mathrm{~m} \end{aligned} \\ &\text { or } \lambda=600 \mathrm{~nm} \end{aligned} $$

Hence, the correct answer is option: 'C'.

Relevant Questions