Problem 1. Light of 5000 A is incident on a circular hole of radius (i) 1 cm and (ii) 1 mm. How many half period zones are contained in the circle if the screen is placed at a distance of 1 m to observe the diffraction?
Hint: (i) Area of the circular hole = πr2 = π cm2
Area of each half period zone= πbλ = 5 x 10-5 x π cm2
No. of half period zones= π/ 5 x 10-5 = 20,000
(ii) Area of circular hole= π x (0.)2 cm2
No. of half period zones= π x (0.1)2 /5 x10-5 x π = 0.2 x 103= 200
Problem 2. Find the radius of first zone is a zone plate of focal length 25 cm for light of wavelength 5000 A
Hint : fn r2n /nλ =f1 =r12/λ (for first zone n=1)
R1 = √f1 λ =0.0354
Problem 3. The spectral lines of sodium D1 and D2 have wavelengths of approximately 5890 A and 5896 A. A sodium lamp sends incident plane wave onto a slit of width 2 Jim. A screen is located 2 m from the slit. Find the spacing between the first maxima of the two sodium lines as measured on the screen.
Hint Distance of the secondary maximum from the centre of the screen is given by
Sin θ=x/D =3/2 λ/a =x=3/2 .Dλ/a
For the two wavelength x1 =3/2 ,Dλ/A and x2 =Dλ2A
Spacing (x2-x1)= 3D /2a (λ2-λ1) =0.9 mm.
Problem 4. In a double slit Fraunhofer diffraction pattern, the screen is 160 em away from the slits. The slit width are 0.08 mm and they are 0.4 mm apart. Calculate the wavelength of light if the fringe spacing is 0.25 em. Also find missing orders.
Hint: The fringe spacing β= λD/ 2d = λ=β(2d) /D=6250 x 10 -8
For missing order a+b /a =n/m
B=0.04 cm and a =0.008 cm
N =6m, where m =1,2,3,4… =n max
Problem 5. A plane transmission grating has 6000 lines/cm. Calculate the highest order spectrum which can be seen with a light of wavelengths 4000 A.
HINT : (a+b) sin θ max =n max λ
N max =(a+b) /λ=4.
Problem 6. The number of lines in a grating X is N 1 and the number of lines in a grating Y is N2, N1 is less than N2. The total ruled width of the two gratings X and Y is same. Compare their resolving powers.
Solution: Resolving power of a grating is given by R.P. = λ/dλ = nN
According to grating equation (a+ b) sin θ nλ or n (a+b) sin θ/λ
From equations (i) and (ii)
R.P. = N(a +b) sin θ /λ
But W = N (a + b), where W = width of the ruled surface
R.P =W sin θ /λ
From equation (iii), it is clear that the resolving power of both the gratings is same and it is independent of the number of lines for a given width of ruled surface.
Problem 7. What other spectral lines in the range 4000 A will coincide with the fifth order line of 6000 A in a grating spectrum.
Hint: (a + b) sin θ = nλ = 5 x 6000 A =5 x 6000A/ n
For n = 6 , λ = 5000 A, and for n = 7 λ = 4285.7 A.
Problem 8. A diffraction grating which has 4000 lines per em is used at normal incidence. Calculate the dispersive power of the grating in the third spectrum in the wavelength region 5000 A.
Hint: dθ/dλ =n /(a+b) cos θ
Also (a+b) sin θ =nλ = sin θ =nλ /a+b
Cos θ =√ (1-sin 2θ) ,then D.P.dθ/dλ =15,000.
Problem 9. What is the minimum number of lines per cm in a 2.5 cm wide grating which will just resolve the two sodium lines (5890 A and 5896 A> in the second order spectrum.
Hint: R.P. λ/dλ = nN = N = λ/dλ =491
Number of lines per em = 491 /2.5 = 196.
Problem 10. How many orders will be visible if the wavelength of the incident radiation is 5000 A and the number of lines on the grating is 2620 in one inch.
Hint: (a+b) sin θ =nλ (a+b) =2.54/2620 cm.
N= (a+b) /λ [as θ =π/2]