Irodov Solutions: Diffraction of Light- 1 Notes | EduRev

I. E. Irodov Solutions for Physics Class 11 & Class 12

JEE : Irodov Solutions: Diffraction of Light- 1 Notes | EduRev

The document Irodov Solutions: Diffraction of Light- 1 Notes | EduRev is a part of the JEE Course I. E. Irodov Solutions for Physics Class 11 & Class 12.
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Q.97. A plane light wave falls normally on a diaphragm with round aperture opening the first N Fresnel zones for a point P on a screen located at a distance b from the diaphragm. The wavelength of light is equal to 2n,. Find the intensity of light /0  in front of the diaphragm if the distribution of intensity of light I (r) on the screen is known. Here r is the distance from the point P. 

Ans.

Irodov Solutions: Diffraction of Light- 1 Notes | EduRev

Irodov Solutions: Diffraction of Light- 1 Notes | EduRev

Q.98. A point source of light with wavelength λ = 0.50 p,m is located at a distance a = 100 cm in front of a diaphragm with round aperture of radius r = 1.0 mm. Find the distance b between the diaphragm and the observation point for which the number of Fresnel zones in the aperture equals k = 3. 

Ans. By definition

Irodov Solutions: Diffraction of Light- 1 Notes | EduRev

for the periphery of the kth zone. Then

Irodov Solutions: Diffraction of Light- 1 Notes | EduRev

 

Q.99. A diaphragm with round aperture, whose radius r can be varied during the experiment, is placed between a point source of light and a screen. The distances from the diaphragm to the source and the screen are equal to a = 100 cm and b = 125 cm. Determine the wavelength of light if the intensity maximum at the centre of the diffraction pattern of the screen is observed at r1 = 1.00 mm and the next maximum at r = 1.29 mm

Ans. Suppose maximum intensity is obtained when the aperture contains k zones. Then a minimum will be obtained for k + 1 zones. Another maximum will be obtained for k + 2 zones. Hence

Irodov Solutions: Diffraction of Light- 1 Notes | EduRev

ThusIrodov Solutions: Diffraction of Light- 1 Notes | EduRev

On putting the values.

 

Q.100. A plane monochromatic light wave with intensity l0  falls normally on an opaque screen with a round aperture. What is the intensity of light I behind the screen at the point for which the aperture (a) is equal to the first Fresnel zone; to the internal half of the first zone; (b) was made equal to the first Fresnel zone and then half of it was closed (along the diameter)? 

Ans. (a) When the aperture is equal to the first Fresnel Zone

 The amplitude is Ax and should be compared with the amplitude y when the aperture is very wide. If I0 is the intensity in the second case the intensity in the first case will be 4lo
When the aperture is equal to the internal half of the first zone Suppose Ain and Aout are the amplitudes due to the two halves o f the first Fresnel zone. Ain and Aout differ in phase by Irodov Solutions: Diffraction of Light- 1 Notes | EduRev because only half the Fresnel zone in involved. Also in magnitude

Irodov Solutions: Diffraction of Light- 1 Notes | EduRev then

Irodov Solutions: Diffraction of Light- 1 Notes | EduRev

Hence following the aigument of the first case.Irodov Solutions: Diffraction of Light- 1 Notes | EduRev

(b) The aperture was made equal to the first Fresnel zone and then half of it was closed along a diameter. In this case the amplitude of vibration is Irodov Solutions: Diffraction of Light- 1 Notes | EduRev Thus

I = l0 .

 

Q.101. A plane monochromatic light wave with intensity Iofalls normally on an opaque disc closing the first Fresnel zone for the observation point P. What did the intensity of light I at the point P become equal to after (a) half of the disc (along the diameter) was removed; (b) half of the external half of the first Fresnel zone was removed (along the diameter)? 

Ans. (a) Suppose the disc does not obstruct light at all. Then 

Irodov Solutions: Diffraction of Light- 1 Notes | EduRev

(because the disc covers the first Fresnel zone only).

Irodov Solutions: Diffraction of Light- 1 Notes | EduRev

Hence the amplitude when half of the disc is removed along a diameter

Irodov Solutions: Diffraction of Light- 1 Notes | EduRev

 (b) In this case

 We write         Irodov Solutions: Diffraction of Light- 1 Notes | EduRev

where Irodov Solutions: Diffraction of Light- 1 Notes | EduRevThe factor i takes account of the Irodov Solutions: Diffraction of Light- 1 Notes | EduRev phase difference between two halves of the first Fresnel zone. Thus

On the other hand Irodov Solutions: Diffraction of Light- 1 Notes | EduRev

so Irodov Solutions: Diffraction of Light- 1 Notes | EduRev

 

Q.102. A plane monochromatic light wave with intensity /0 falls normally on the surfaces of the opaque screens shown in Fig. 5.20. Find the intensity of light I at a point P 

 Irodov Solutions: Diffraction of Light- 1 Notes | EduRev 

(a) located behind the corner points of screens 1-3 and behind the edge of half-plane 4;
 (b) for which the rounded-off edge of screens 5-8 coincides with the boundary of the first Fresnel zone. Derive the general formula describing the results obtained for screens 1-4; the same, for screens 5-8. 

Ans. W hen the screen is fully transparent, the. amplitude of vibrations isIrodov Solutions: Diffraction of Light- 1 Notes | EduRev (with intensity

Irodov Solutions: Diffraction of Light- 1 Notes | EduRev

Irodov Solutions: Diffraction of Light- 1 Notes | EduRev

(2) In this case Irodov Solutions: Diffraction of Light- 1 Notes | EduRev of the plane is blacked out so

Irodov Solutions: Diffraction of Light- 1 Notes | EduRev

Irodov Solutions: Diffraction of Light- 1 Notes | EduRev

Irodov Solutions: Diffraction of Light- 1 Notes | EduRev

Irodov Solutions: Diffraction of Light- 1 Notes | EduRev

In 5 to 8 the first term in the expression for the amplitude is the contribution of the plane part and the second term gives the expression for the Fresnel zone part. In general in (5) to

Irodov Solutions: Diffraction of Light- 1 Notes | EduRev is the angle covered by the screen.

 

Q.103. A plane light wave with wavelength λ = 0.60 1.t.m falls normally on a sufficiently large glass plate having a round recess on the opposite side (Fig. 5.21). For the observation point P that recess corresponds to the first one and a half Fresnel zones. Find the depth h of the recess at which the intensity of light at the point P is (a) maximum; (b) minimum; (c) equal to the intensity of incident light. 

Irodov Solutions: Diffraction of Light- 1 Notes | EduRev

Ans. We would require the contribution to the amplitude of a wave at a point from half a Fresnel zone. For this we proceed directly from the Fresnel Huyghens principle. The complex amplitude is written as

Irodov Solutions: Diffraction of Light- 1 Notes | EduRev

Irodov Solutions: Diffraction of Light- 1 Notes | EduRev

Irodov Solutions: Diffraction of Light- 1 Notes | EduRev

Irodov Solutions: Diffraction of Light- 1 Notes | EduRev

For the first Fresnel zone Irodov Solutions: Diffraction of Light- 1 Notes | EduRev

Thus

Irodov Solutions: Diffraction of Light- 1 Notes | EduRev

For the next half zone

Irodov Solutions: Diffraction of Light- 1 Notes | EduRev

If we calculate the contribution of the full 2nd Fresnel zone we will get - Av If we take account of the factors K (φ) and Irodov Solutions: Diffraction of Light- 1 Notes | EduRevwhich decrease monotonically we exp ect the contribution to change to - A2. Thus we write for the contribution of the half zones in the 2nd Fresnel zone as

Irodov Solutions: Diffraction of Light- 1 Notes | EduRev

The part lying in the recess has an extra phase difference equal to Irodov Solutions: Diffraction of Light- 1 Notes | EduRev Thus the full amplitude is (note that the correct form is Irodov Solutions: Diffraction of Light- 1 Notes | EduRev

Irodov Solutions: Diffraction of Light- 1 Notes | EduRev

Irodov Solutions: Diffraction of Light- 1 Notes | EduRev

(a) For maximum intensity sin

Irodov Solutions: Diffraction of Light- 1 Notes | EduRev

Irodov Solutions: Diffraction of Light- 1 Notes | EduRev

(b) For minimum intensity

Irodov Solutions: Diffraction of Light- 1 Notes | EduRev

 

Irodov Solutions: Diffraction of Light- 1 Notes | EduRev

Irodov Solutions: Diffraction of Light- 1 Notes | EduRev

 

Q.104. A plane light wave with wavelength  λ  and intensity Ifalls normally on a large glass plate whose opposite side serves as an opaque screen with a round aperture equal to the first Fresnel zone for the observation point P. In the middle of the aperture there is a round recess equal to half the Fresnel zone. What must the depth h of that recess be for the intensity of light at the point P to be the highest? What is this intensity equal to?

Ans. The contribution to the wave amplitude of the inner half-zone is 

Irodov Solutions: Diffraction of Light- 1 Notes | EduRev

Irodov Solutions: Diffraction of Light- 1 Notes | EduRev

With phase factor this becom es Irodov Solutions: Diffraction of Light- 1 Notes | EduRevThe contribution of

Irodov Solutions: Diffraction of Light- 1 Notes | EduRev

and the intensity is

Irodov Solutions: Diffraction of Light- 1 Notes | EduRev

Here Irodov Solutions: Diffraction of Light- 1 Notes | EduRev .s the intensity of the incident light which is the same as the intensity due to an aperture of infinite extent (and no recess). Now

l is maximum when Irodov Solutions: Diffraction of Light- 1 Notes | EduRev

 Irodov Solutions: Diffraction of Light- 1 Notes | EduRev

Irodov Solutions: Diffraction of Light- 1 Notes | EduRev

 

Q.105. A plane light wave with wavelength λ = 0.57 um falls normally on a surface of a glass (n = 1.60) disc which shuts one and a half Fresnel zones for the observation point P. What must the minimum thickness of that disc be for the intensity of light at the point P to be the highest? Take into account the interference of light on its passing through the disc. 

Ans.  We follow Ihe argument of 5.103. we find that the contribution of the first Fresnel zone is 

Irodov Solutions: Diffraction of Light- 1 Notes | EduRev

For the n ext h a lf zone it i  Irodov Solutions: Diffraction of Light- 1 Notes | EduRev

(The contribution of the remaining part of the 2nd Fresnel zone will be  Irodov Solutions: Diffraction of Light- 1 Notes | EduRev

If the disc has a thickness h, the extra phase difference suffered by the light wave in passing through' the disc will be

Irodov Solutions: Diffraction of Light- 1 Notes | EduRev

Thus the amplitude at P will be

Irodov Solutions: Diffraction of Light- 1 Notes | EduRev

The corresponding intensity will be

Irodov Solutions: Diffraction of Light- 1 Notes | EduRev

 The intensity will be a maximum when

or   Irodov Solutions: Diffraction of Light- 1 Notes | EduRev

i.e Irodov Solutions: Diffraction of Light- 1 Notes | EduRev

Irodov Solutions: Diffraction of Light- 1 Notes | EduRev

 

Q.106. A plane light wave with wavelength λ = 0.54 um goes through a thin converging lens with focal length f = 50 cm and an aperture stop fixed immediately after the lens, and reaches a screen placed at a distance b = 75 cm from the aperture stop. At what aperture radii has the centre of the diffraction pattern on the screen the maximum illuminance? 

Ans. Here the focal point acts as a virtual source of light. This means that we can take spherical waves conveiging towards F. Let us divide these waves intofresnel zones just after they emerge from the stop. We write 
Irodov Solutions: Diffraction of Light- 1 Notes | EduRev

Irodov Solutions: Diffraction of Light- 1 Notes | EduRev

Here r is the radius of the mth fresnel zone and A is the distance to the left of the foot of the perpendicular. Thus 

Irodov Solutions: Diffraction of Light- 1 Notes | EduRev

The intensity maxima are observed when an odd number of Fresnel zones are exposed by the stop. Thus

Irodov Solutions: Diffraction of Light- 1 Notes | EduRev

 

Q.107. A plane monochromatic light wave falls normally on a round aperture. At a distance b = 9.0 m from it there is a screen showing a certain diffraction pattern. The aperture diameter was decreased η = 3.0 times. Find the new distance b' at which the screen should be positioned to obtain the diffraction pattern similar to the previous one but diminished η times. 

Ans. or the radius of the periphery of the kth zone we have

Irodov Solutions: Diffraction of Light- 1 Notes | EduRev

If the aperture diameter is reduced η times it will produce a similar deftraction pattern (reduced η times) if the radii of the Fresnel zones are also η times less. Thus

Irodov Solutions: Diffraction of Light- 1 Notes | EduRev

 

Q.108. An opaque ball of diameter D = 40 mm is placed between a source of light with wavelength λ = 0.55 um and a photographic plate. The distance between the source and the ball is equal to a = 12 m and that between the ball and the photographic plate is equal to b = 18 m. Find:
 (a) the image dimension y' on the plate if the transverse dimension of the source is y = 6.0 mm;
 (b) the minimum height of irregularities, covering the surface of the ball at random, at which the ball obstructs light. 

Note. As calculations and experience show, that happens when the height of irregularities is comparable  with the width of the Fresnel zone along which the edge of an opaque screen passes

Ans. (a) If a point source is placecd before an opaque ball, the diffraction pattern consists of a bright spot inside a dark disc followed by fringes. The bright spot is on the line joining the point source and the centre of the ball. When the object is a finite source of transverse diamensiony, every point of the source has its corresponding image on the line joining that point and the centre of the ball. Thus the transverse dimension of the image is given by

Irodov Solutions: Diffraction of Light- 1 Notes | EduRev

 (b) The minimum height of the irregularities covering the surface of the ball at random, at which the ball obstructs light is^according to the note at the end of the problem, corn* parable with the width of the Fresnel zone along which the edge of opaque screen passes. SO

Irodov Solutions: Diffraction of Light- 1 Notes | EduRev

Tofind Δr we note that

Irodov Solutions: Diffraction of Light- 1 Notes | EduRev

Where D = diameter of the disc (= diameter of the last Fresnel zone) and Δk = 1 Thus 

Irodov Solutions: Diffraction of Light- 1 Notes | EduRev

 

Q.109. A point source of monochromatic light is positioned in front of a zone plate at a distance a = 1.5 m from it. The image of the source is formed at a distance b = 1.0 m from the plate. Find the focal length of the zone plate. 

Irodov Solutions: Diffraction of Light- 1 Notes | EduRev

Ans.

Irodov Solutions: Diffraction of Light- 1 Notes | EduRev

Irodov Solutions: Diffraction of Light- 1 Notes | EduRev

Irodov Solutions: Diffraction of Light- 1 Notes | EduRev

This is the principle focal length. Other maxima are obtained when

Irodov Solutions: Diffraction of Light- 1 Notes | EduRev

These focal lengths are alsoIrodov Solutions: Diffraction of Light- 1 Notes | EduRev

 

Q.110. A plane light wave with wave- length λ = 0.60 um and intensity I, falls normally on a large glass plate whose side view is shown in Fig. 5.22. At what height h of the ledge will the intensity of light at points located directly below be
 (a) minimum;
 (b) twice as low as I0  (the losses due to reflection are to be neglected). 

Ans. Just below the edge the amplitude of the wave is given by  

Irodov Solutions: Diffraction of Light- 1 Notes | EduRev

Irodov Solutions: Diffraction of Light- 1 Notes | EduRev

Here the quantity in the brackets is the contribution of various Fresnel zones; the factor Irodov Solutions: Diffraction of Light- 1 Notes | EduRev is to take account of the division of the plate into two parts by the ledge; the phase factor 5 is given by

Irodov Solutions: Diffraction of Light- 1 Notes | EduRev

and takes into account the extra length traversed by the waves on the left.

Using  Irodov Solutions: Diffraction of Light- 1 Notes | EduRev

we get  Irodov Solutions: Diffraction of Light- 1 Notes | EduRev

and the corresponding intensity is

Irodov Solutions: Diffraction of Light- 1 Notes | EduRev

Irodov Solutions: Diffraction of Light- 1 Notes | EduRev

Irodov Solutions: Diffraction of Light- 1 Notes | EduRev

 

Q.111. A plane monochromatic light wave falls normally on an opaque half-plane. A screen is located at a distance b = 100 cm behind the half-plane. Making use of the Cornu spiral (Fig. 5.19), find: (a) the ratio of intensities of the first maximum and the neighbouring minimum; (b) the wavelength of light if the first two maxima are separated by a distance Δx = 0.63 mm. 

Ans.  (a) From the Cornu’s spiral, the intensity of t^e first maximum is given as

Irodov Solutions: Diffraction of Light- 1 Notes | EduRev

and the intensity of the first minimum is given by

Irodov Solutions: Diffraction of Light- 1 Notes | EduRev

 so the required ratio is

Irodov Solutions: Diffraction of Light- 1 Notes | EduRev

(b) The value of the distance x is related to the parameter v in Fresnel's integral by

Irodov Solutions: Diffraction of Light- 1 Notes | EduRev

For the first two maxima the distances x1 , x 2 are related to the parameters v1, v2 by

Irodov Solutions: Diffraction of Light- 1 Notes | EduRev

or   Irodov Solutions: Diffraction of Light- 1 Notes | EduRev

From the Cornu’s spiral the positions of the maxima are 

Irodov Solutions: Diffraction of Light- 1 Notes | EduRev

 

Q.112. A plane light wave with wavelength 0.60 pm falls normally on a long opaque strip 0.70 mm wide. Behind it a screen is placed at a distance 100 cm. Using Fig. 5.19, find the ratio of intensities of light in the middle of the diffraction pattern and at the edge of the geometrical shadow.

Ans. We shall use the equation written down in 5.103, the Fresnel-Huyghens formula. 

Irodov Solutions: Diffraction of Light- 1 Notes | EduRev

Suppose we want tofind the intensity at P which is such that the coordinates of the edges (x-coordinates) with respect to P are xand - x1. Then, the amplitude at P is

Irodov Solutions: Diffraction of Light- 1 Notes | EduRev (1)

Irodov Solutions: Diffraction of Light- 1 Notes | EduRev

Irodov Solutions: Diffraction of Light- 1 Notes | EduRev

Irodov Solutions: Diffraction of Light- 1 Notes | EduRev

Irodov Solutions: Diffraction of Light- 1 Notes | EduRev

Irodov Solutions: Diffraction of Light- 1 Notes | EduRev

where

Irodov Solutions: Diffraction of Light- 1 Notes | EduRev

 Rough evaluation of the integrals using cornu’s spiral gives

Irodov Solutions: Diffraction of Light- 1 Notes | EduRev

 

Q.113. A plane monochromatic light wave falls normally on a long rectangular slit behind which a screen is positioned at a distance b = 60 cm. First the width of the slit was adjusted so that in the middle of the diffraction pattern the lowest minimum was observed. After widening the slit by Δh = 0.70 mm, the next minimum was obtained in the centre of the pattern. Find the wavelength of light. 

Ans. If the aperture has width A then the parameters ( v , - v ) associated w ith

Irodov Solutions: Diffraction of Light- 1 Notes | EduRev

The intensity of light at O on the screen is obtained as the square of the amplitude A of the wave at O which is

Irodov Solutions: Diffraction of Light- 1 Notes | EduRev

Irodov Solutions: Diffraction of Light- 1 Notes | EduRev

Irodov Solutions: Diffraction of Light- 1 Notes | EduRev

By definition v corresponds to the first minimum of the intensity. This means

when we increase Irodov Solutions: Diffraction of Light- 1 Notes | EduRev relates to the second

Irodov Solutions: Diffraction of Light- 1 Notes | EduRev

 

Q.114. A plane light wave with wavelength λ = 0.65 p.m falls normally on a large glass plate whose opposite side has a long rectangular recess 0.60 mm wide. Using Fig. 5.19, find the depth h of the recess at which the diffraction pattern on the screen 77 cm away from the plate has the maximum illuminance at its centre. 

Ans. Let a = width of the recess and 

Irodov Solutions: Diffraction of Light- 1 Notes | EduRev

be die parameter along Cornu’s spiral corresponding to the half-width of the recess. The amplitude of the diffracted wave is given by

Irodov Solutions: Diffraction of Light- 1 Notes | EduRev

is the extra phase due to the recess. (Actually an extra phase e -lδ appears outside the recess. When we take it out and absorb it in the constant we get the expression written).
Thus the amplitude is

Irodov Solutions: Diffraction of Light- 1 Notes | EduRev

Irodov Solutions: Diffraction of Light- 1 Notes | EduRev

Irodov Solutions: Diffraction of Light- 1 Notes | EduRev

We write

Irodov Solutions: Diffraction of Light- 1 Notes | EduRev

For maximum intensity

or Irodov Solutions: Diffraction of Light- 1 Notes | EduRev

so  Irodov Solutions: Diffraction of Light- 1 Notes | EduRev

 

Q.115. A plane light wave with wavelength λ = 0.65 p.m falls normally on a large glass plate whose opposite side has a ledge and an opaque strip of width a = 0.30 mm (Fig. 5.23). A screen is placed at a distance b = 110 cm from the plate. The height h of the ledge is such that the intensity of light at point 2 of the

 Irodov Solutions: Diffraction of Light- 1 Notes | EduRev

screen is the highest possible. Making use of Fig. 5.19, find the ratio of intensities at points 1 and 2. 

Ans.

Irodov Solutions: Diffraction of Light- 1 Notes | EduRev

Using the method of problem 5.103 we can immediately write down the amplitudes at 1 and 2. We get :

At 1Irodov Solutions: Diffraction of Light- 1 Notes | EduRev

At 2Irodov Solutions: Diffraction of Light- 1 Notes | EduRev

where Irodov Solutions: Diffraction of Light- 1 Notes | EduRev

is the parameter of Cornu’s spiral and constant factor is common to 1 and 2. With the usual notation

Irodov Solutions: Diffraction of Light- 1 Notes | EduRev

and the result  Irodov Solutions: Diffraction of Light- 1 Notes | EduRev

We find the ratio of intensities as

Irodov Solutions: Diffraction of Light- 1 Notes | EduRev

 (The constants in A1 and A2 must be the same by symmetry) 

Irodov Solutions: Diffraction of Light- 1 Notes | EduRev

Irodov Solutions: Diffraction of Light- 1 Notes | EduRev

Irodov Solutions: Diffraction of Light- 1 Notes | EduRev

Irodov Solutions: Diffraction of Light- 1 Notes | EduRev

 

Q.116. A plane monochromatic light wave of intensity Io  falls normally on an opaque screen with a long slit having a semicircular 

Irodov Solutions: Diffraction of Light- 1 Notes | EduRev

cut on one side (Fig. 5.24). The edge of the cut coincides with the boundary line of the first Fresnel zone for the observation point P. The width of the slit measures 0.90 of the radius of the cut. Using Fig. 5.19, find the intensity of light at the point P.

Ans.  Irodov Solutions: Diffraction of Light- 1 Notes | EduRev

The contribution of the full 1st Fresnel zone has been evaluated in 5.103. The contribution of the semi-circle is one half of it and is

Irodov Solutions: Diffraction of Light- 1 Notes | EduRev

Irodov Solutions: Diffraction of Light- 1 Notes | EduRev

Thus the contribution of the slit is

Irodov Solutions: Diffraction of Light- 1 Notes | EduRev

Thus the intensity at the observation point P on the screeii is

Irodov Solutions: Diffraction of Light- 1 Notes | EduRev

Irodov Solutions: Diffraction of Light- 1 Notes | EduRev

 

Q.117. A plane monochromatic light wave falls normally on an opaque screen with a long slit whose shape is shown in Fig. 5.25. Making use of Fig. 5.19, find the ratio of intensities of light at points 1, 2, and 3 located behind the screen at equal distances from it. For point 3 the rounded-off edge of the slit coincides with the boundary line of the first Fresnel zone.

Ans. From the statement of the problem we know that the width of the slit = diameter of the first Fresnel Irodov Solutions: Diffraction of Light- 1 Notes | EduRev where b is the distance of the observation point from the slit

 We calculate the amplitudes by evaluating the integral of problem 5.103 We get

Irodov Solutions: Diffraction of Light- 1 Notes | EduRev

 

Irodov Solutions: Diffraction of Light- 1 Notes | EduRev

Irodov Solutions: Diffraction of Light- 1 Notes | EduRev

Irodov Solutions: Diffraction of Light- 1 Notes | EduRev

where the contribution of the 1st half Fresnel zone (in A3, first term) has been obtained from the last problem.
Thus

Irodov Solutions: Diffraction of Light- 1 Notes | EduRev

So    Irodov Solutions: Diffraction of Light- 1 Notes | EduRev

ThusIrodov Solutions: Diffraction of Light- 1 Notes | EduRev

 

Q.118. A plane monochromatic light wave falls normally on an opaque screen shaped as a long strip with a round hole in the middle. For the observation point P the hole corresponds to half the Fresnel zone, with the hole diameter being η = 1.07 times less than the width of the strip. Using Fig. 5.19, find the intensity of light at the point P provided that the intensity of the incident light is equal to I0

Ans. 

Irodov Solutions: Diffraction of Light- 1 Notes | EduRev

Irodov Solutions: Diffraction of Light- 1 Notes | EduRev

We use 

Irodov Solutions: Diffraction of Light- 1 Notes | EduRev

Irodov Solutions: Diffraction of Light- 1 Notes | EduRev

Thus

Irodov Solutions: Diffraction of Light- 1 Notes | EduRev

where we have used

Irodov Solutions: Diffraction of Light- 1 Notes | EduRev

Thus the intensity is

Irodov Solutions: Diffraction of Light- 1 Notes | EduRev

 From Cornu’s Spiral,

Irodov Solutions: Diffraction of Light- 1 Notes | EduRev

As before

Irodov Solutions: Diffraction of Light- 1 Notes | EduRev

 

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