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Light of λ = 550 nm is incident as parallel beam on a slit of width 0.1 m . Find the angular width and linear width of the principal maximum in the diffraction pattern on a screen at a distance of 1.1 m from thw slit. Which of these widths will not change if the screen were moved to a distance of 2.2 m from the slit?

User UncleBob
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Final answer:

Answer is 2.2 m. The angular width of the principal maximum in the diffraction pattern can be calculated using the formula θ = λ / w, where θ is the angular width, λ is the wavelength of light, and w is the width of the slit. The linear width of the principal maximum can be calculated using the formula D = λL / w, where D is the linear width, L is the distance between the screen and the slit, and w is the width of the slit. If the screen is moved to a different distance, the angular width of the principal maximum will remain the same, but the linear width will change.

Step-by-step explanation:

The angular width of the principal maximum (first minimum) in the diffraction pattern on a screen can be calculated using the formula:

θ = λ / w

where θ is the angular width, λ is the wavelength of light, and w is the width of the slit.

Given λ = 550 nm and w = 0.1 mm = 0.0001 m:

θ = (550 nm) / (0.0001 m)

= 5.5 x 10-3 radians

The linear width of the principal maximum can be calculated using the formula:

D = λL / w

where D is the linear width, L is the distance between the screen and the slit, and w is the width of the slit.

Given λ = 550 nm,

w = 0.1 mm

= 0.0001 m, and L = 1.1 m:

D = (550 nm) * (1.1 m) / (0.0001 m)

= 6050 nm =

6.05 μm

If the screen were moved to a distance of 2.2 m from the slit, the angular width of the principal maximum would remain the same, but the linear width would change.

Therefore, the linear width of the principal maximum will not change if the screen is moved to a distance of 2.2 m from the slit.

User The Ref
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