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Class 11-science H C VERMA Solutions Physics Chapter 17: Light Waves

Light Waves Exercise 380

Solution 1

  

  

 Hz

  

 Hz

So, Range of frequency

  

Solution 2

(a)   

  

 Hz

(b)   

  

 nm

(c)  [frequency depends only upon source, not an medium]

 Hz

(d)  

  

 m/s

Solution 3

  

For   

  

 m/s

For   

  

 m/s 

Solution 4

  

  

 1.25 

Solution 5

(a)   

  

D=1m

  

  

  

 0.05mm

(b)   

  

  

  

 0.5mm

Light Waves Exercise 381

Solution 6

  

  

  

 m

λ = 400 nm

Solution 7

  

(a) distance=  

  

   

=0.25 mm

(b) Fringe width= mm

No. of fringes obtained in 1 cm= =20 fringes.

Solution 8

  

Distance between consecutive bright

=  

   

 m

=1.47 mm

Solution 9

Angular fringe width,   

  

  

  

  

Solution 10

  

Distance of first maxima from center=  

Linear separation between first maxima,   

  

 mm

Solution 11

Central maxima for both wavelength forms at same position at centre of screen.

Let mth maxima of violet coincides with   maxima of red light.

  

  

  

  

Least value of m is 7. 

Solution 12

  

  

  

Solution 13

(a) Change in optical path=  

(b)   

  

For minimum thickness, D-1 will be formed.

  

  

Solution 14

  

  

  

  

  

Hence, 14.5 fringes will shift.

Solution 15

Distance moved by fringes after introduction of slab

  

  

Now, distance of screen is doubled, new fringe width

  

  

  

Given,

  

  

  

  

 589.2 nm 

Solution 16

(a)   

  

 m

(b) No. of fringes shifted

Path difference due to sheets,

  

  

  

 m

  

  

  

25 Fringes will be shifted completely

 

  

 

Position of first maxima on one side= 0.43β 

  

  

=0.021 cm

and on other side=0.57β 

  

 0.028 cm

Solution 17

Path difference at centre of screen

  

  

Minima is formed at the centre of screen 

  

  

For minimum thickness, n=1

  

Solution 18

 

(a)   

  

  

  

(b) Initially path difference at center of screen=  

  

  

  

  

  

So, B-15 is formed.

Now, when identical paper piece is pasted, path difference at center will be zero. So, B-0 will be formed.

Hence, 15 fringes will shift.

Solution 19

Fringe width

  

  

 m

 =0.9 mm

Solution 20

  

 

Path difference between waves meeting at center of screen will be due to pre-slit path difference.

  

  

  

So, dark fringe is obtained at centre of screen.  

Solution 21

(a) The phase of a light wave reflecting from plane mirror will change by π.

At point just above O, the  and  

  

  

So, minima is formed i.e., zero intensity.

 

 

(b) Let y distance above O, maxima is obtained at point P.

  

  

  

 For maxima,   

  

 Put n=1 for   

  

  

Light Waves Exercise 382

Solution 22

Fringe width,

  

  

 0.35 mm 

Solution 23

Intensity of light from source   

Intensity of light after reflection   

  

  

  

Solution 24

Distance between slits=d

Distance between apparent sources and screen

=  

=  

Fringe width   

  

Solution 25

(a) The wavelength which forms dark (minima) at hole will be absent.

  

  

  

 nm

For n=1, λ=2000 nm

For n=2, λ=667 nm

For n=3, λ=400 nm

Thus, 667 nm and 400 nm are absent in the range of (400 nm-700 nm).

(b) The wavelength which forms bright (maxima) will have strong intensity.

  

  

  

  

  (in nm)

For n=1; λ=1000 nm

For n=2; λ=500 nm

For n=3; λ=334 nm

So, in Range of (400-700 nm) , light of wavelength 500 nm will have maxima. 

Solution 26

  

(a) For minima at O

  

  

  

  

 By binomial expansion

  

  

For minima d, n=1

  

(b) At point O, D-1 is formed

 

 

Bright fringe next to D-1 will be B-0.

So, point at which path difference is zero B-0 will be formed.

By symmetry the point is in front of slit B.

So, minimum distance x=d.

(c) distance between B-0 and D-1 is d

  

  

Solution 27

Let point P at distance x will have minimum intensity i.e., D-2

Path difference=  

  

  

  

  

  

Solution 28

(a) Path difference   

  

  

By binomial expansion

  

  

  

(b) Path difference between wave from slit A and C is

  

  

  

By binomial expansion

  

  

  

  

  

  

  

  

  

  

By superposition principle

 

  

  

  

Thus, the resultant intensity is three times the intensity due to the individual slits.

Solution 29

  

For point at position, y=0.5cm

Path difference=  

  

 m

Phase difference,   

  

  

  

  

  

  

  

Solution 30

(a)   

  

  

  

  

  

  

  

(b)   

  

  

  

  

  

  

  

Solution 31

  

  

  

  

  

  

  

  

  

 m

Solution 32

Distance of the point from central maxima where intensity falls by half.

  

  

  

  

  

  

  

  

Line-width= 2y

  

  

Light Waves Exercise 383

Solution 33

(a)   

At β/2 distance from central maxima, minima will be obtained.

So, intensity from S4=0

  

(b)  

At  distance from central maxima, first order bright will be obtained.

So, intensity from   

  

(c)   

  

  

  

So, intensity at   

Now,

  

Solution 34

(a)  

Distance of slit   and   from perpendicular bisector is y= .

Path difference from  and  wave at  = .

Similarly, at   path difference=  

i.e., dark fringes are formed at   and  .

So, intensity of light at   and   are zero.

Hence, intensity at P is also zero.

(b)   

Distance of slit   and   from perpendicular bisector is y= .

Path difference from  and  wave at  = .

Similarly, at   path difference=  

Phase difference at   and   will be=  

When  , i.e., distance from perpendicular bisector is   

  

  intensity was I at point P.

At phase difference  , intensity at point P is I.

(c)   

Distance of slit   and   from perpendicular bisector is y= .

So, at   and  , bright fringe will be formed.

Intensity at P is 2I.

Solution 35

Minima is obtained in reflected light

  

  

  

 n

For refractive index between 1.2 and 1.5, put n=5

  

 1.32 

Solution 36

Maxima is obtained in reflected light.

  

for   , put π=0

  

  

 100 nm 

Solution 37

Maxima in reflected light

  

  

  

For n=4, λ=666 nm

For n=5, λ=532 nm

For n=6, λ=443 nm 

Solution 38

Condition for maxima in reflected light

  

  

  

 nm

For n=3, λ=714 nm

For n=4, λ=556 nm

For n=5, λ=455 nm  

Solution 39

For minima, in diffraction

  

  

λ=2.5 cm 

Solution 40

  

  

 m

R=0.683 cm

Diameter of central bright spot=2R

=2(0.683)

=1.37 cm 

Solution 41

  

  

 m

Diameter of the central bright spot,  m.