A beam of light consisting of two wavelengths \( 5000\,\mathring{\mathrm{A}} \) and \( 6000\,\mathring{\mathrm{A}} \) is used to obtain interference fringes in Young's double slit experiment. The least distance from the central maxima, where the bright fringes due to both wavelengths coincide, will be: (If separation between slits = 1 mm and separation between slits and screen is 1 m)
Answer & explanation
Correct answer: option 2
The correct answer is Option (2) → \( 3\,\mathrm{mm} \)
In Young's double slit experiment the \( n \)th bright fringe is at \( y_n = n\lambda D/d \). Two bright fringes coincide where the \( n_1 \)th fringe of \( \lambda_1 \) falls on the \( n_2 \)th fringe of \( \lambda_2 \).
\( \displaystyle \frac{n_1 \lambda_1 D}{d} = \frac{n_2 \lambda_2 D}{d} \;\Rightarrow\; n_1 \lambda_1 = n_2 \lambda_2 \)
\( \displaystyle \frac{n_1}{n_2} = \frac{\lambda_2}{\lambda_1} = \frac{6000}{5000} = \frac{6}{5} \)
The smallest whole numbers are \( n_1 = 6 \) (for \( 5000\,\mathring{\mathrm{A}} \)) and \( n_2 = 5 \) (for \( 6000\,\mathring{\mathrm{A}} \)).
\( \lambda_1 = 5000\,\mathring{\mathrm{A}} = 5\times10^{-7}\,\mathrm{m} \)
\( d = 1\,\mathrm{mm} = 10^{-3}\,\mathrm{m} \)
\( D = 1\,\mathrm{m} \)
\( \displaystyle y = \frac{n_1 \lambda_1 D}{d} = \frac{6 \times 5\times10^{-7} \times 1}{10^{-3}} = 3\times10^{-3}\,\mathrm{m} = 3\,\mathrm{mm} \)
The same value comes from the other wavelength: \( 5 \times 6\times10^{-7} \times 1/10^{-3} = 3\times10^{-3}\,\mathrm{m} \).
Note: NTA picked this question from an NCERT (2026-27) exercise and changed the figures. Do not ignore NCERT while practising numericals.
NCERT (2026-27):
Exercises – …
