Question

In a diffraction grating experiment, light of 600 nm wavelength produces a first-order maximum 0.350 mm from the central maximum on a distant screen. A second monochromatic source produces a third-order maximum 0.870 mm from the central maximum when it passes through the same diffraction grating. What is the wavelength of the light from the second source

Answer 1

Answer:

497.143 nm.

Explanation:

Diffraction grating experiment is actually done by passing light through diffraction glasses, the passage of the light causes some patterns which can be seen on the screen. This is because light is a wave and it can spread.

The solution to the question is through the use of the formula in the equation (1) below;

Sin θ = m × λ. ---------------------------------(1).

Where m takes values from 0, 1, 2, ...(that is the diffraction grating principal maxima).

Also, m × λ = dc/ B -------------------------------------------(2).

We are to find the second wavelength, therefore;

λ2 =( m1/c1) × (c2/m2) × λ1 ------------------------(3).

Where c1 and c2 are the order maximum and m = order numbers. Hence;

λ2 = (1/ .350) × (.870/3) × 600 = 497.143 nm.

Answer 2

Answer:

∧2 = 497 nm

Explanation:

d = the spacing between every two lines

D = the distance from the grating to the screen

∧1 = wavelength from the fist source

∧2 = wavelength from the second source

The first order maximum y1 = 0.350 mm

The third order maximum y2 = 0.870 mm

∧1 = 600 nm

∧2 = ?

Therefore,

1 × ∧1 = d × y1 / D  

3 × ∧2 = d × y2 / D

divide the ∧2 by ∧1

3∧2/∧1 = y2/y1

∧2 = (y2 × ∧1 )/ 3 × y1

∧2 = (0.870 × 600) / 3 × 0.350

∧2 = 522/1.05

∧2 = 497.142857143

∧2 = 497 nm