Chapter 36: Q6P (page 1109)

Monochromatic light of wavelength $441 nm$ is incident on a narrow slit. On a screen $2.00 m$ away, the distance between the second diffraction minimum and the central maximum is $1.50 cm$ . (a) Calculate the angle of diffraction role="math" localid="1663145961694" $θ$ of the second minimum. (b) Find the width of the slit.

Short Answer

Expert verified

(a) The angle of diffraction for second order minima is $0.429 °$ .

(b) The width of slit is $1.18 \times 10^{- 4} m$ .

Step by step solution

Identification of given data

The wavelength of the monochromatic light is $λ = 441 nm$ .

The distance of screen from narrow slit is $D = 2 m$ .

The distance between central minima and second order minima is $w = 1.50 cm$ .

The order of second minima is $m = 2$ .

Concept used

The diffraction is the phenomenon in which the light bends on the edges of the slit and pattern of light is formed on a screen.

Determination of angle of diffraction for second order minima

(a)

The angle of diffraction for second order minima is given as:

$\tan θ = \frac{w}{D}$

Substitute all the values in equation.

$\tan θ = \frac{(1.50 cm) (\frac{1 m}{100 cm})}{2 m} θ = 0.429 °$

Therefore, the angle of diffraction for second order minima is $0.429 °$ .

Determination of width of slit

(b)

The width of slit is given as:

$m λ = d \sin θ$

Substitute all the values in equation.

$(2) (441 nm) = d (\sin 0.429 °) d = 1.18 \times 10^{- 4} m$

Therefore, the width of slit is $1.18 \times 10^{- 4} m$ .

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Short Answer

Step by step solution

Identification of given data

Concept used

Determination of angle of diffraction for second order minima

Determination of width of slit

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