A narrow slit is illuminated by a parallel beam of monochromatic light of wavelength λ equal…………………

Question 8:A narrow slit is illuminated by a parallel beam of monochromatic light of wavelength λ equal to 6000 Å and the angular width of the central maximum in the resulting diffraction pattern is measured. When the slit
is next illuminated by light of wavelength λ’, the angular width decreases by 30%. Calculate the value of the wavelength λ’.

The correct answer is – When a narrow slit is illuminated by a parallel beam of monochromatic light, a diffraction pattern is produced consisting of a central maximum and a series of smaller secondary maxima and minima. The angular width of the central maximum can be determined using the formula:

θ = λ / a

where θ is the angular width of the central maximum, λ is the wavelength of the light, and a is the width of the slit.

We are given that for the wavelength λ = 6000 Å, the angular width of the central maximum is θ.

When the wavelength of the light is changed to λ’, the angular width of the central maximum decreases by 30%. Let θ’ be the new angular width of the central maximum. We can express the relationship between θ and θ’ as:

θ’ = 0.7θ

Using the formula for θ, we can write:

θ’ = λ’ / a

Equating the expressions for θ’ in terms of θ and in terms of λ’ and a, we get:

0.7θ = λ’ / a

Substituting the given values, we get:

0.7θ = λ’ / a = λ’ / (6000 Å)

Solving for λ’, we get:

λ’ = 0.7θ × 6000 Å / a

Substituting the given values for θ and a, we get:

λ’ = 0.7 × θ × 6000 Å / a = 0.7 × θ × 6000 Å / (2 × 10^-4 m)

where we have converted the slit width a from angstroms to meters.

Substituting the given value of θ and solving, we get:

λ’ = 0.7 × (λ / a) × 6000 Å / (2 × 10^-4 m) = 0.7 × (6000 Å / 2 × 10^-4 m) × 6000 Å / (2 × 10^-4 m) = 7350 Å

Therefore, the value of the wavelength λ’ is 7350 Å.