Question

What is the wavelength of light falling on double slits separated by 2.00μm if the third-order maximum is at an angle of 60.0º?

a) 0.95 μm
b) 1.15 μm
c) 1.00 μm
d) 0.75 μm

Answer 1

The wavelength of light falling on double slits separated by 2.00 µm if the third-order maximum is at an angle of 60.0° is 1.15 µm.

Step-by-step explanation:

To find the wavelength of light falling on double slits separated by 2.00 µm with the third-order maximum at an angle of 60.0°, we utilize the formula for constructive interference in a double-slit experiment:

d sin(θ) = mλ, where:

• d is the distance between the slits,
• θ is the angle of the maximum,
• m is the order of the maximum (the integer)
• λ is the wavelength.

The given values are:

• d = 2.00 µm,
• θ = 60.0°,
• m = 3 (for third-order maximum).

Plugging these values into the formula gives us:

2.00 µm × sin(60.0°) = 3λ

Calculate sin(60.0°), which is √3/2:

2.00 µm × (√3/2) = 3λ

λ = (2.00 µm × √3/2) / 3

λ = 1.15 µm

Thus, the wavelength of the light is 1.15 µm, corresponding to option b).