A thin slit illuminated by light of frequency f produces its first dark band at ± 38.2 ∘ in air. When the entire apparatus (slit, screen and space in between) is immersed in an unknown transparent liquid, the slit's first dark bands occur instead at ± 17.3 ∘.

Find the refractive index of the liquid.

the refractive index of the liquid is 2.08

Explanation:

The diffraction formula is:

d sin (θ) = mλ

where

d is the diffraction m = 1, 2 λ is the wavelength of the light

Thus,

d sin (θ₁) = m λ₁ (1) and

d sin (θ₂) = m λ₂ (2)

Dividing equation (1) by equation (2) you get

sin (θ₁) / sin (θ₂) = λ₁ / λ₂ (3)

λ can be expressed as

λ = v / f

where

v is the speed of light f is the frequency of light

Therefore,

λ₁ = v₁ / f

λ₂ = v₂ / f

Thus,

λ₁ / λ₂ = v₁ / v₂

Therefore, substituting the above expression for λ₁ / λ₂ into equation (3), we get

sin (θ₁) / sin (θ₂) = v₁ / v₂

where

v₁ is the speed of light in air

The expression for the refractive index is

n = c / v, which can also be expressed as

n = v₁ / v₂

Therefore,

n = sin (θ₁) / sin (θ₂)

n = sin (38.2°) / sin (17.3°)

n = 2.08

Therefore, the refractive index of the liquid is 2.08