Question 368: Waves Optics

Question

In Young's double slit experiment, the $$10^{	ext {th }}$$ maximum of wavelength '$$\lambda_1$$' is at a distance of '$$Y_1$$' from the central maximum. When the wavelength of the source is changed to '$$\lambda_2$$', $$5^{th}$$ maximum is at a distance '$$Y_2$$' from the central maximum. The ratio $$\frac{Y_1}{Y_2}$$ is

Accepted Answer

In Young's double slit experiment, the position of the $$n^{	ext{th}}$$ maximum for a given wavelength can be calculated using the formula:
$$Y = \frac{n \lambda D}{d}$$
where:
$$Y$$ is the distance from the central maximum,
$$n$$ is the order of the maximum,
$$\lambda$$ is the wavelength of the light,
$$D$$ is the distance between the slits and the screen,
$$d$$ is the distance between the two slits.
According to the given problem:
The $$10^{	ext{th}}$$ maximum for wavelength $$\lambda_1$$ is at a distance $$Y_1$$ from the central maximum:
$$Y_1 = \frac{10 \lambda_1 D}{d}$$
The $$5^{	ext{th}}$$ maximum for wavelength $$\lambda_2$$ is at a distance $$Y_2$$ from the central maximum:
$$Y_2 = \frac{5 \lambda_2 D}{d}$$
We need to find the ratio $$\frac{Y_1}{Y_2}$$:
$$\frac{Y_1}{Y_2} = \frac{\frac{10 \lambda_1 D}{d}}{\frac{5 \lambda_2 D}{d}} = \frac{10 \lambda_1}{5 \lambda_2} = \frac{2 \lambda_1}{\lambda_2}$$
Therefore, the correct answer is:
Option A
$$\frac{2 \lambda_1}{\lambda_2}$$

Answer

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

Answer

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

Answer

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

Answer

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

Waves Optics

In Young's double slit experiment, the $10^{\text {th }} maximum of wavelength '\lambda_1' is at a distance of 'Y_1' from the central maximum. When the wavelength of the source is changed to '\lambda_2', 5^{th} maximum is at a distance 'Y_2' from the central maximum. The ratio \frac{Y_1}{Y_2}$ is

Choose the correct answer: