Fraunhofer diffraction from a single-slit of width 1 . 24 × 10 - 6 m is observed with light of wavelength 6200 Å . Calculate the angular width of the central maximum.

Given, Wavelength of light, λ = 6200 Å = 6200 × 10 - 10 m Width of the slit, d = 1 . 24 × 10 - 6 m We can use the relationship d sin θ = λ to find the angular position of the first minima. Where θ is the angular position of the first minima. sin θ = λ d = 6200 × 10 - 10 1 . 24 × 10 - 6 = 1 2 ⇒ θ = 30 ° So, angular width of the central maxima = 2 θ = 60 °

E M B I B E

ISC Physics Class 12, Part-2>Chapter 8 - Diffraction of Light>QUESTIONS>Q 4

MEDIUM

12th ICSE

IMPORTANT

Earn 100

Draw a labelled graph to show variation in intensity of diffracted light with angular position in a single-slit diffraction experiment.

Important Questions on Diffraction of Light

HARD

12th ICSE

IMPORTANT

ISC Physics Class 12, Part-2>Chapter 8 - Diffraction of Light>QUESTIONS>Q 1

Define diffraction of light and name the wavefront used in Fraunhofer diffraction. Obtain the relation

e \sin θ = λ

for the first minima.

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12th ICSE

IMPORTANT

ISC Physics Class 12, Part-2>Chapter 8 - Diffraction of Light>QUESTIONS>Q 2

Draw diagram of diffraction of light due to a single slit and draw a graph showing the variation of intensity of light in a single slit Fraunhofer diffraction.

HARD

12th ICSE

IMPORTANT

ISC Physics Class 12, Part-2>Chapter 8 - Diffraction of Light>QUESTIONS>Q 3

Compare interference and diffraction. Obtain the expression for half and full angular width of central maxima in diffraction pattern.

MEDIUM

12th ICSE

IMPORTANT

ISC Physics Class 12, Part-2>Chapter 8 - Diffraction of Light>QUESTIONS>Q 1

Fraunhofer diffraction from a single-slit of width