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GATE 2023 PH – Question 54

Optical Physics · Interference: spatial and temporal coherence, dielectric films, Newton’s ring, multiple-beam interference, Michelson interferometer, Fabry-Perot interferometer and etalon · 2 marks · Multiple choice

Different spectral lines of the Balmer series (transitions n→2, with n being the principal quantum number) fall one at a time on a Young’s double slit apparatus. The separation between the slits is d and the screen is placed at a constant distance from the slits. What factor should d be multiplied by to maintain a constant fringe width for various lines, as n takes different allowed values?

  1. $(n^2-4)/(4n^2)$
  2. $(n^2+4)/(4n^2)$
  3. $4n^2/(n^2-4)$
  4. $4n^2/(n^2+4)$

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Correct answer: (C) $4n^2/(n^2-4)$

Explanation

**Fringe width** in a double-slit experiment is
$$\beta=\frac{\lambda D}{d},$$
where $D$ is the distance to the screen. For the fringe width to stay constant while $\lambda$ changes, the slit separation must be proportional to the wavelength: $d\propto\lambda$.

**Balmer series wavelengths:**
$$\frac1\lambda=R\left(\frac1{2^2}-\frac1{n^2}\right)=R\,\frac{n^2-4}{4n^2}\;\Rightarrow\;\lambda=\frac{4n^2}{R\,(n^2-4)} .$$

So $d$ must be multiplied by a factor proportional to $\lambda$:
$$\frac{4n^2}{n^2-4}\quad(\text{option C}).$$