Question 335: Waves Optics

Question

In Young's double slit experiment, the fifth maximum with wavelength '$$\lambda_1$$' is at a distance '$$y_1$$' and the same maximum with wavelength '$$\lambda_2$$' is at a distance '$$y_2$$' measured from the central bright band. Then $$\frac{y_1}{y_2}$$ is equal to [D and $$d$$ are constant]

Accepted Answer

The equations for the position of the fringe from the central maxima are given as
$$\begin{aligned}
\mathrm{y}_1 & =\frac{5 \lambda_1 \mathrm{D}}{\mathrm{d}} \
\mathrm{y}_2 & =\frac{5 \lambda_2 \mathrm{D}}{\mathrm{d}} \
	herefore \quad \frac{\mathrm{y}_1}{\mathrm{y}_2} & =\frac{\lambda_1}{\lambda_2}
\end{aligned}$$

Answer

$$\frac{\lambda_1}{\lambda_2}$$

Answer

$$\frac{\lambda_2}{\lambda_1}$$

Answer

$$\frac{\lambda_1^2}{\lambda_2^2}$$

Answer

$$\frac{\lambda_2^2}{\lambda_1^2}$$

Waves Optics

In Young's double slit experiment, the fifth maximum with wavelength '$\lambda_1' is at a distance 'y_1' and the same maximum with wavelength '\lambda_2' is at a distance 'y_2' measured from the central bright band. Then \frac{y_1}{y_2} is equal to [D and d$ are constant]

Choose the correct answer: