In a single slit diffraction experiment, the aperture of the slit is 3 mm. A monochromatic light of wavelength 620 nm is normally incident on the slit. If the distance between the slit and the screen is 1.2 m, then the separation between the first order minima and the third order maxima on one side of the screen is
Answer & explanation
Correct answer: option 3
The correct answer is Option (3) → \( 6.20 \times 10^{-4}\,\mathrm{m} \)
In single slit diffraction the minima on the screen are at \( y = n\lambda D/a \), and the secondary maxima lie roughly midway between them, at \( y = (n + 1/2)\,\lambda D/a \).
\( a = 3\,\mathrm{mm} = 3\times10^{-3}\,\mathrm{m} \)
\( \lambda = 620\,\mathrm{nm} = 6.2\times10^{-7}\,\mathrm{m} \)
\( D = 1.2\,\mathrm{m} \)
\( \displaystyle \frac{\lambda D}{a} = \frac{6.2\times10^{-7} \times 1.2}{3\times10^{-3}} = 2.48\times10^{-4}\,\mathrm{m} \)
The first minimum (\( n = 1 \)) and the third maximum (\( n = 3 \)) are at
\( \displaystyle y_{1,\min} = 1 \times \frac{\lambda D}{a} \qquad y_{3,\max} = 3.5 \times \frac{\lambda D}{a} \)
\( \displaystyle \Delta y = (3.5 - 1)\,\frac{\lambda D}{a} = 2.5 \times 2.48\times10^{-4} = 6.20\times10^{-4}\,\mathrm{m} \)