Question 33: Waves Optics

Question

In the Young's double slit experiment $$n^{	ext {th }}$$ bright for red coincides with $$(n+1)^{	ext {th }}$$ bright for violet. Then the value of '$$n$$' is: (given: wave length of red light $$=6300^{\circ} \mathrm{A}$$ and wave length of violet $$=4200^{\circ} \mathrm{A}$$).

Accepted Answer

In Young's double slit experiment, the position of the $$n^{	ext{th}}$$ bright fringe is given by the equation:
$$ x_n = \frac{n \lambda D}{d} $$
where:
$$x_n$$ is the position of the fringe
$$n$$ is the order number of the fringe
$$\lambda$$ is the wavelength of the light used
$$D$$ is the distance between the slits and the screen
$$d$$ is the distance between the two slits
In this problem, we are given that the $$n^{	ext{th}}$$ bright fringe for red light coincides with the $$(n+1)^{	ext{th}}$$ bright fringe for violet light. This means:
$$ n \lambda_{	ext{red}} = (n + 1) \lambda_{	ext{violet}} $$
Given the wavelengths:
$$\lambda_{	ext{red}} = 6300^{\circ} 	ext{A}$$
$$\lambda_{	ext{violet}} = 4200^{\circ} 	ext{A}$$
We can substitute these values into the equation:
$$ n \cdot 6300 = (n + 1) \cdot 4200 $$
Simplifying this equation:
$$ 6300n = 4200n + 4200 $$
$$ 2100n = 4200 $$
$$ n = 2 $$
So, the value of '$$n$$' is 2.
Thus, the correct option is:
Option A: 2

Answer

2

Answer

4

Answer

3

Answer

1

Waves Optics

In the Young's double slit experiment $n^{\text {th }} bright for red coincides with (n+1)^{\text {th }} bright for violet. Then the value of 'n' is: (given: wave length of red light =6300^{\circ} \mathrm{A} and wave length of violet =4200^{\circ} \mathrm{A}$).

Choose the correct answer: