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GATE 2023 PH – Question 44

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

Young’s double slit experiment is performed using a beam of C60 (fullerene) molecules, each molecule being made up of 60 carbon atoms. When the slit separation is 50 nm, fringes are formed on a screen kept at a distance of 1 m from the slits. Now, the experiment is repeated with C70 molecules with a slit separation of 92.5 nm. The kinetic energies of both the beams are the same. The position of the 4th bright fringe for C60 will correspond to the nth bright fringe for C70. What is the value of n (rounded off to the nearest integer) ?

  1. 5
  2. 6
  3. 7
  4. 8

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Correct answer: (D) 8

Explanation

In a double-slit experiment the bright fringes are at $y_n=\dfrac{n\lambda D}{d}$, where $D$ is the screen distance and $d$ the slit separation.

**de Broglie wavelength** at the same kinetic energy: $\lambda=\dfrac{h}{\sqrt{2mK}}\propto\dfrac1{\sqrt m}$. The mass is proportional to the number of carbon atoms, so
$$\frac{\lambda_{70}}{\lambda_{60}}=\sqrt{\frac{60}{70}}.$$

**Same position:** the 4th bright fringe of C₆₀ ($d=50$ nm) and the $n$th bright fringe of C₇₀ ($d=92.5$ nm) at the same $y$:
$$\frac{4\lambda_{60}D}{50}=\frac{n\lambda_{70}D}{92.5}\;\Rightarrow\;n=4\cdot\frac{92.5}{50}\cdot\frac{\lambda_{60}}{\lambda_{70}}=4\times1.85\times\sqrt{\frac{70}{60}}=7.40\times1.0801=7.99 .$$

The value is $n\approx\mathbf{8}$ (option D).