Two slits are illuminated with green light (λ = 540 nm). The slits are 0.05 mm apart and the distance to the screen is 1.5 m. At what distance (in mm) from the central maximum on the screen is the average intensity 50% of the intensity of the central maximum?

0.21486 mm

Explanation:

The formula for the maximum intensity is given by;

I = I_o•cos²(Φ/2)

Now,we are not given Φ but it can be expressed in terms of what we are given as; Φ = πdy/(λL)

Where;

y is the distance from the central maximum

d is the distance between the slits

λ is the wavelength

L is the distance to the screen

Thus;

I = I_o•πdy/(λL)

We are given;

d = 0.05 mm = 0.5 × 10^(-3) m

λ = 540 nm = 540 × 10^(-9) m

L = 1.25 m

I/I_o = 50% = 0.5

From earlier, we saw that;

I = I_o•πdy/(λL)

We have I/I_o = 0.5

Thus;

I/I_o = πdy/(λL)

Plugging in the relevant values;

0.5 = (π × 0.5 × 10^(-3) × y)/(540 × 10^(-9) × 1.25)

Making y the subject, we have;

y = (0.5 × 540 × 10^(-9) × 1.25)/(π × 0.5 × 10^(-3))

y = 0.00021486 m

Converting to mm, we have;

y = 0.21486 mm

the answer would be a