CBSE 2025, Set 1

Question

In Young's double slit experiment, the separation between the two slits is 1·0 mm and the screen is 1·0 m away from the slits. A beam of light consisting of two wavelengths 500 nm and 600 nm is used to obtain interference fringes. Calculate :

(a) the distance between the first maxima for the two wavelengths.

(b) the least distance from the central maximum, where the bright fringes due to both the wavelengths coincide.

Edvaya · Accepted Answer

The correct answer is Subjective. Given: $d = 1\cdot0$ mm $= 10^{-3}$ m, $D = 1\cdot0$ m, $\lambda_1 = 500$ nm, $\lambda_2 = 600$ nm The position of the $n$th bright fringe: $y_n = \dfrac{n\lambda D}{d}$ **(a) Distance between first maxima for the two wavelengths:** $$y_1(\lambda_1) = \frac{1 	imes 500 	imes 10^{-9} 	imes 1}{10^{-3}} = 0\cdot5 	imes 10^{-3} 	ext{ m} = 0\cdot5 	ext{ mm}$$ $$y_1(\lambda_2) = \frac{1 	imes 600 	imes 10^{-9} 	imes 1}{10^{-3}} = 0\cdot6 	imes 10^{-3} 	ext{ m} = 0\cdot6 	ext{ mm}$$ $$\boxed{\Delta y = y_1(\lambda_2) - y_1(\lambda_1) = 0\cdot1 	ext{ mm} = 1 	imes 10^{-4} 	ext{ m}}$$ **(b) Least distance from central maximum where bright fringes coincide:** For coincidence: $n_1\lambda_1 = n_2\lambda_2$ $$\frac{n_1}{n_2} = \frac{\lambda_2}{\lambda_1} = \frac{600}{500} = \frac{6}{5}$$ The minimum integers satisfying this are $n_1 = 6$ and $n_2 = 5$. $$y = \frac{n_1\lambda_1 D}{d} = \frac{6 	imes 500 	imes 10^{-9} 	imes 1}{10^{-3}}$$ $$\boxed{y = 3 	imes 10^{-3} 	ext{ m} = 3 	ext{ mm}}$$ **Verification:** $n_2\lambda_2 D/d = 5 	imes 600 	imes 10^{-9}/10^{-3} = 3 	imes 10^{-3}$ m ✓ > **Tip:** To find coincidence of bright fringes, find the LCM of the fringe widths: LCM(500, 600) = 3000 nm → coincides at 3000 nm path difference → $y = 3\lambda_1 D/d = 3 	imes 0.5$ mm = 1.5 mm × 2 = 3 mm.

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