Q144 / 155 · WBJEE 2019 · Physics

In Young's experiment for the interference of light, the separation between the slits is $d$ and the distance of the screen from the slits is $D$. If $D$ is increased by $0.5 \%$ and $d$ is decreased by $0.3 \%$ then for the light of a given wavelength, which one of the following is true? "The fringe width.........

  • Correct answer
    A
    increases by $0.8 \%$
  • B
    decreases by $0.8 \%$
  • C
    increases by $0.2 \%$
  • D
    decreases by $0.2 \%$

Solution

As we know that, fringe width, $\beta=\frac{\lambda D}{d}$ where, $\lambda=$ wavelength of light $D=$ Distance of the screen from the slits $d=$ separation between the slits Now, $\frac{\Delta \beta}{\beta} \times 100=\frac{\Delta D}{D} \times 100-\frac{\Delta d}{d} \times 100$ According to the question. $$ \frac{\Delta \beta}{\beta} \times 100=0.5-(-0.3)=0.8 \% $$ Hence, the fringe width increases by $0.2 \%$

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

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