Draw a neat labelled diagram of Young ' s Double Slit experiment. Show that β = λ α , where the terms have their usual meaning ( either for bright or dark fringe ) .

Consider a point P on the viewing screen, the screen is located at a perpendicular distance D from the screen containing the slits S 1 and S 2 , which are separated by a distance d as shown in figure. Let us assume that the source of light is monochromatic having a wavelength λ . Under these conditions, the waves emerging from S 1 and S 2 have the same frequency and amplitude and are in phase. The light intensity on the screen at P is the resultant of the lights coming from both the slits. As seen in the diagram a wave from the lower slit travels farther than a wave from the upper slit. This difference in path is called the path difference. Therefore, if ' x ' is the path difference between the waves from slits S 2 and S 1 at point P , then x = r 2 - r 1 = d sin θ ‐ equation 1 where it is assumed that r 1 and r 2 are parallel, which is approximately true because D is much greater than d . The value of this path difference determines whether or not the two waves are in phase, when they arrive at point P . x = d sin θ = n λ ‐ equation 2 In order to find the position of maxima measured vertically from O to P i.e., y we assume that D > > d . Under these conditions, θ is small, and so the approximation sin θ ~ tan θ . Therefore, from ∆ P O Q in figure, we find that sin θ ≈ tan θ = y D ‐ equation 3 Substituting this in equation 2 , we have d y D = n λ ‐ equation 4 Rewriting the above equation, we have y = D n λ d This gives the distance of the n t h maxima from the centre O of the screen. For n = 0 , 1 , 2 , 3 , 4 , . . . the values of y are y 0 = 0 . . . position of central bright fringe y 1 = D λ d . . . position of first bright fringe y 2 = 2 D λ d . . . position of second bright fringe y n = n D λ d . . . position of the n t h bright fringe The spacing between any two bright fringes gives the width of the dark fringe. Therefore, y n - y n - 1 = n D λ d - n - 1 D λ d = D λ d This gives the width of the dark fringe. So the fringe width is given by, β = D λ d

Draw a neat labelled diagram of Young ' s Double Slit experiment. Show that β = λ α , where the terms have their usual meaning ( either for bright or dark fringe ) .

Consider a point P on the viewing screen, the screen is located at a perpendicular distance D from the screen containing the slits S 1 and S 2 , which are separated by a distance d as shown in figure. Let us assume that the source of light is monochromatic having a wavelength λ . Under these conditions, the waves emerging from S 1 and S 2 have the same frequency and amplitude and are in phase. The light intensity on the screen at P is the resultant of the lights coming from both the slits. As seen in the diagram a wave from the lower slit travels farther than a wave from the upper slit. This difference in path is called the path difference. Therefore, if ' x ' is the path difference between the waves from slits S 2 and S 1 at point P , then x = r 2 - r 1 = d sin θ ‐ equation 1 where it is assumed that r 1 and r 2 are parallel, which is approximately true because D is much greater than d . The value of this path difference determines whether or not the two waves are in phase, when they arrive at point P . x = d sin θ = n λ ‐ equation 2 In order to find the position of maxima measured vertically from O to P i.e., y we assume that D > > d . Under these conditions, θ is small, and so the approximation sin θ ~ tan θ . Therefore, from ∆ P O Q in figure, we find that sin θ ≈ tan θ = y D ‐ equation 3 Substituting this in equation 2 , we have d y D = n λ ‐ equation 4 Rewriting the above equation, we have y = D n λ d This gives the distance of the n t h maxima from the centre O of the screen. For n = 0 , 1 , 2 , 3 , 4 , . . . the values of y are y 0 = 0 . . . position of central bright fringe y 1 = D λ d . . . position of first bright fringe y 2 = 2 D λ d . . . position of second bright fringe y n = n D λ d . . . position of the n t h bright fringe The spacing between any two bright fringes gives the width of the dark fringe. Therefore, y n - y n - 1 = n D λ d - n - 1 D λ d = D λ d This gives the width of the dark fringe. So the fringe width is given by, β = D λ d

E M B I B E

EMBIBE CHAPTER WISE PREVIOUS YEAR PAPERS FOR PHYSICS>Chapter 21 - Diffraction of Light>ICSE-2018>Q 1

MEDIUM

12th ICSE

IMPORTANT

Earn 100

In Young’s double slit experiment, show graphically how intensity of light varies with distance.

Important Questions on Diffraction of Light

EASY

12th ICSE

IMPORTANT

EMBIBE CHAPTER WISE PREVIOUS YEAR PAPERS FOR PHYSICS>Chapter 21 - Diffraction of Light>ICSE-2018>Q 2

In Fraunhofer diffraction, how is the angular width of the central bright fringe affected when slit separation is increased ?

MEDIUM

12th ICSE

IMPORTANT

EMBIBE CHAPTER WISE PREVIOUS YEAR PAPERS FOR PHYSICS>Chapter 21 - Diffraction of Light>ICSE-2019>Q 3

What should be the path difference between two waves reaching a point for obtaining constructive interference in Young $'$ s Double Slit experiment?

HARD

12th ICSE

IMPORTANT

EMBIBE CHAPTER WISE PREVIOUS YEAR PAPERS FOR PHYSICS>Chapter 21 - Diffraction of Light>ICSE-2019>Q 4

Draw a neat labelled diagram of Young $'$ s Double Slit experiment. Show that

β = \frac{λ}{α}

, where the terms have their usual meaning

(

either for bright or dark fringe

)

.

MEDIUM

12th ICSE

IMPORTANT

EMBIBE CHAPTER WISE PREVIOUS YEAR PAPERS FOR PHYSICS>Chapter 21 - Diffraction of Light>ICSE-2020>Q 5

In Young's double-slit experiment, the screen is kept at a distance of $1.2 m$ from the plane of the slits. The two slits are separated by $5 mm$ and illuminated with monochromatic light having wavelength $600 nm$ . Calculate the fringe width i.e. fringe separation of the interference pattern.

MEDIUM

12th ICSE

IMPORTANT

EMBIBE CHAPTER WISE PREVIOUS YEAR PAPERS FOR PHYSICS>Chapter 21 - Diffraction of Light>ICSE-2020>Q 6

In Young's double-slit experiment, the screen is kept at a distance of $1.2 m$ from the plane of the slits. The two slits are separated by $5 mm$ and illuminated with monochromatic light having wavelength $600 nm$ . Calculate the distance of $10^{t h}$ bright fringe from the centre of the pattern.

MEDIUM

12th ICSE

IMPORTANT

EMBIBE CHAPTER WISE PREVIOUS YEAR PAPERS FOR PHYSICS>Chapter 21 - Diffraction of Light>ICSE-2020>Q 7

Draw a labelled graph of the intensity of diffracted light $(I)$ versus angle $(θ)$ in the Fraunhofer diffraction experiment for a single-slit diffraction.

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