Q130 / 180 · NEET 2013 (Karnataka) · Physics

In Young's double slit experiment the distance between the slits and the screen is doubled. The separation between the slits reduced to half. As a result the fringe width:

  • A
    is halved
  • Correct answer
    B
    becomes four times
  • C
    remains unchanged
  • D
    is doubled

Solution

In Young's double slit experiment fringe width $$ \beta=\frac{\lambda D}{d} $$ where $D=$ distance between slit and screen $d=$ distance between slit According to the question $D$ is double $\rightarrow 2 D$ and $d$ is half $\rightarrow$ $\frac{d}{2}$ Ther New fringe wilth will be $\frac{\lambda \times 2 D}{\frac{d}{2}}=\frac{4 \lambda D}{d}=4\left(\frac{\lambda D}{d}\right)=\beta$ $\beta^1=4 \beta$ i.e., becomes four times

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

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