Question

In a Young’s double slit experiment,

λ=500 nm, d=1·0 mm and D=1·0 m. Find the minimum distance from the central maximum for which the intensity is half of the maximum intensity.

Answer 1

Given :

D=1m

d= 1m= 10⁻³ m

l= 500nm= 500 x 10⁻⁹m= 500 x 10⁻⁷m

For intensity to be half the maximum intensity :

y= λD/4d

= 5 x 10⁻⁷ x 1/ 4 x 10⁻⁷

y= 1.25 x 10⁻⁴ m

∴The minimum distance from the central maximum for which the intensity is half of the maximum intensity is 1.25 x 10⁻⁴ m

Answer 2

Answer:

The minimum distance from the central maximum which the intensity is half of the maximum intensity is 0.125 mm.

Explanation:

Given that,

Wavelength = 500 nm

Distance = 1.0 m

Width = 1.0 mm

We need to calculate the minimum distance from the central maximum which the intensity is half of the maximum intensity

Using formula of distance

$y=\dfrac{\lambda D}{4d}$

Put the value into the formula

$y=\dfrac{500\times10^{-9}\times1.0}{4\times1\times10^{-3}}$

$y=0.125\ mm$

Hence, The minimum distance from the central maximum which the intensity is half of the maximum intensity is 0.125 mm.