Question 321: Waves Optics

Question

In Young's double slit experiment, $$8^{	ext {th }}$$ maximum with wavelength '$$\lambda_1$$' is at a distance '$$d_1$$' from the central maximum and $$6^{	ext {th }}$$ maximum with wavelength '$$\lambda_2$$' is at a distance '$$\mathrm{d}_2$$'. Then $$\frac{\mathrm{d}_2}{\mathrm{~d}_1}$$ is

Accepted Answer

$$\begin{aligned}
& \mathrm{d} \propto \mathrm{n} \lambda \
	herefore \quad & \frac{\mathrm{d}_2}{\mathrm{~d}_1}=\frac{\mathrm{n}_2 \lambda_2}{\mathrm{n}_1 \lambda_1}=\frac{6 \lambda_2}{8 \lambda_1} \
	herefore \quad & \frac{\mathrm{d}_2}{\mathrm{~d}_1}=\frac{3 \lambda_2}{4 \lambda_1}
\end{aligned}$$

Answer

$$\frac{3 \lambda_1}{4 \lambda_2}$$

Answer

$$\frac{3 \lambda_2}{4 \lambda_1}$$

Answer

$$\frac{4 \lambda_1}{3 \lambda_2}$$

Answer

$$\frac{4 \lambda_2}{3 \lambda_1}$$

Waves Optics

In Young's double slit experiment, $8^{\text {th }} maximum with wavelength '\lambda_1' is at a distance 'd_1' from the central maximum and 6^{\text {th }} maximum with wavelength '\lambda_2' is at a distance '\mathrm{d}_2'. Then \frac{\mathrm{d}_2}{\mathrm{~d}_1}$ is

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