CBSE 2025, Compartment, Set 1

Question

Light of wavelength $640	ext{ nm}$ incident on the slits in a Young's double-slit experiment produces an interference pattern of fringe width $7.2	ext{ mm}$. Calculate the wavelength of light which will produce a pattern of fringe width $8.1	ext{ mm}$.

Edvaya · Accepted Answer

The correct answer is Subjective. **SOLUTION** The formula for the fringe width ($\beta$) in Young's double-slit experiment is given by: $$\beta = \frac{\lambda D}{d}$$ Since the distance to the screen ($D$) and the distance between the slits ($d$) remain completely identical across both experimental cases, the fringe width is directly proportional to the wavelength: $$\beta \propto \lambda$$ $$\implies \frac{\beta_1}{\beta_2} = \frac{\lambda_1}{\lambda_2}$$ Given: * $\lambda_1 = 640	ext{ nm}$ * $\beta_1 = 7.2	ext{ mm}$ * $\beta_2 = 8.1	ext{ mm}$ Rearranging the formula to find the new wavelength ($\lambda_2$): $$\lambda_2 = \lambda_1 	imes \left(\frac{\beta_2}{\beta_1}ight)$$ $$\lambda_2 = 640 	imes \left(\frac{8.1}{7.2}ight)$$ $$\lambda_2 = 640 	imes \left(\frac{9}{8}ight)$$ $$\lambda_2 = 80 	imes 9$$ $$\lambda_2 = 	extbf{720 nm}$$

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