Two parallel beams of monochromatic light pass through a long slit of width 0.2 mm one by one. The angular divergence in which most of the light is diffracted for both the beams are \( 2.25\times10^{-3}\,\mathrm{rad} \) and \( 4.50\times10^{-3}\,\mathrm{rad} \) respectively. If the wavelength of the first beam is \( 3000\,\mathring{\mathrm{A}} \), then the wavelength of the second beam is:
Answer & explanation
Correct answer: option 3
The correct answer is Option (3) → \( 6000\,\mathring{\mathrm{A}} \)
Most of the light of a single-slit diffraction pattern lies in the central maximum, whose angular width is \( 2\lambda/a \). For the same slit width \( a \), this angular width is directly proportional to the wavelength.
\( \displaystyle \theta = \frac{2\lambda}{a} \;\Rightarrow\; \frac{\theta_2}{\theta_1} = \frac{\lambda_2}{\lambda_1} \)
\( \displaystyle \lambda_2 = \lambda_1\,\frac{\theta_2}{\theta_1} = 3000 \times \frac{4.50\times10^{-3}}{2.25\times10^{-3}} = 6000\,\mathring{\mathrm{A}} \)
Note: It seems that there is a mistake in the original NTA paper, which prints \( 2.25\times10^{-3}\,\mathrm{rad} \) when it should read \( 3.0\times10^{-3}\,\mathrm{rad} \). Worked directly, \( \lambda_2 = a\theta_2/2 \) would give \( 4500\,\mathring{\mathrm{A}} \); the answer key follows the ratio of the two printed angles, which gives \( 6000\,\mathring{\mathrm{A}} \) (Option 3).