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Solved Examples on Wave Optics Question 1:- A slit of width a is illuminated by white light. (a) For what value of a will the first minimum for red light of λ = 650 nm be at θ = 15°? (b) What is the wavelength λ' of the light whose first side diffraction maximum is at 15°, thus coinciding with the first minimum for the red light? Solution:- (a) At the first minimum, m = 1 in equation [ a sinθ = mλ, for m = 1,2,3,...]. Solving for a, we then find a = mλ / sinθ = (1) (650 nm) / (sin15°) = 2511 nm ≈ 2.5 µm Therefore, the value of a the first minimum for red light of λ = 650 nm be at θ = 15° would be 2.5 µm. For the incident light to flare out that much (±15°) the slit has to be very fine indeed, amounting to about four times the wavelength. Note that a fine human hair may be about 100 µm in diameter. (b) This maximum is about halfway between the first and second minima produced with wavelength λ'. we can find it without too much error by putting m = 1.5 in equation [a sinθ = mλ, for m = 1,2,3,...], obtaining a sinθ = 1.5 λ' Solving for λ' and substituting known data give λ' = a sinθ/1.5 = (2511 nm) (sin15°)/1.5 = 430 nm From the above observation we conclude that, the wavelength λ' of the light whose first side diffraction maximum is at 15° would be 430 nm. Light of this wavelength is violet. the first side maximum for light of wavelength 430 nm will always coincide with the first minimum for light of wavelength 650 nm, no matter what the slit width. If the slit is relatively narrow, the angle θ at which this overlap occurs will be relatively large, and conversely. ____________________________________________________________________________________ Question 2:- A two slit Young's experiment is done with monochromatic light of wavelength 6000 Å. Slits are 2 mm apart and fringes are observed on a screen placed 10 cm away from the slits. If a transparent plate of thickness 0.5 mm is placed in front of one of the slit, interference pattern shifts by 5 mm. What is the refractive index of transparent plate? Solution:- We know that, (µ –1)t = Xd/D Or, µ = 1+Xd/Dt Substituting the value of X, d , D and t in the equation µ = 1+Xd/Dt, we get, Or, µ = 1+ (0.5) (0.2)/(10) (0.05) = 1+0.2 = 1.2 From the above observation we conclude that, the refractive index of transparent plate would be 1.2. _____________________________________________________________________________________________ Question 3:- Angular width of central maximum in the Fraunhoffer's diffraction pattern is measured. Slit is illuminated by the light of another wavelength, angular width decreases by 30%. What is the wavelength of light used? Solution:- For first diffraction min. d sin θ = λ if angle is small, sin θ =θ and dθ =λ That is, half angular width, θ = λ/d Full angular width w = 2θ = 2λ/d Also w' = 2λ'/d λ'/λ = w'/w or λ' =λ (w'/w) = 6000 × 0.7 = 4200 Å From the above observation we conclude that, the wavelength of light used would be 4200 Å. ____________________________________________________________________________________________ Question 4:- A beam of light consisting of two wavelengths 6500A°and 5200A° is used to obtain interference fringes in a Young’s double slit experiment (a) Find the distance of the third fringe on the screen from the central maximum for the wavelength 6500A°. (b) What is the least distance from the central maximum where the bright fringes due to both wavelength coincide? (c) The distance between the slits is 2mm and the distance between the plane of the slits and screen is 120cm. What is the fringe width for l = 6500A°? Solution:- (a) The width of the fringe Dλ/d Then distance of the third fringe 3w = 3Dλ/d = [(3) (120) (6500) (10^{-8})]/[0.2] = 0.117 cm Thus, the distance of the third fringe on the screen from the central maximum for the wavelength 6500A° would be 0.117 cm. (b) Let m^{th} and n^{th} bright fringe of the wavelength coincide. Now position m^{th} bright fringe is y_{m} = mλ_{1} (D/d) and y_{n}_{ }= nλ_{2} (D/d) Or, m/n = λ_{1}/λ_{2 }= 5200/6500 = 4/5 (c) Fringe width, w = λD/d = [6500 × 10^{-10} ×1.2] / [2× 10^{-13}] = 3900× 10^{-7} m = 0.039 cm From the above observation we conclude that, the fringe width for l = 6500A° would be 0.039 cm.
A slit of width a is illuminated by white light.
(a) For what value of a will the first minimum for red light of λ = 650 nm be at θ = 15°?
(b) What is the wavelength λ' of the light whose first side diffraction maximum is at 15°, thus coinciding with the first minimum for the red light?
(a) At the first minimum, m = 1 in equation [ a sinθ = mλ, for m = 1,2,3,...]. Solving for a, we then find
a = mλ / sinθ
= (1) (650 nm) / (sin15°)
= 2511 nm
≈ 2.5 µm
Therefore, the value of a the first minimum for red light of λ = 650 nm be at θ = 15° would be 2.5 µm. For the incident light to flare out that much (±15°) the slit has to be very fine indeed, amounting to about four times the wavelength. Note that a fine human hair may be about 100 µm in diameter.
(b) This maximum is about halfway between the first and second minima produced with wavelength λ'. we can find it without too much error by putting m = 1.5 in equation [a sinθ = mλ, for m = 1,2,3,...], obtaining
a sinθ = 1.5 λ'
Solving for λ' and substituting known data give
λ' = a sinθ/1.5
= (2511 nm) (sin15°)/1.5
= 430 nm
From the above observation we conclude that, the wavelength λ' of the light whose first side diffraction maximum is at 15° would be 430 nm. Light of this wavelength is violet. the first side maximum for light of wavelength 430 nm will always coincide with the first minimum for light of wavelength 650 nm, no matter what the slit width. If the slit is relatively narrow, the angle θ at which this overlap occurs will be relatively large, and conversely.
____________________________________________________________________________________
A two slit Young's experiment is done with monochromatic light of wavelength 6000 Å. Slits are 2 mm apart and fringes are observed on a screen placed 10 cm away from the slits. If a transparent plate of thickness 0.5 mm is placed in front of one of the slit, interference pattern shifts by 5 mm. What is the refractive index of transparent plate?
We know that,
(µ –1)t = Xd/D
Or, µ = 1+Xd/Dt
Substituting the value of X, d , D and t in the equation µ = 1+Xd/Dt, we get,
Or, µ = 1+ (0.5) (0.2)/(10) (0.05)
= 1+0.2
= 1.2
From the above observation we conclude that, the refractive index of transparent plate would be 1.2.
_____________________________________________________________________________________________
Angular width of central maximum in the Fraunhoffer's diffraction pattern is measured. Slit is illuminated by the light of another wavelength, angular width decreases by 30%. What is the wavelength of light used?
For first diffraction min. d sin θ = λ
if angle is small, sin θ =θ and dθ =λ
That is, half angular width, θ = λ/d
Full angular width w = 2θ = 2λ/d
Also w' = 2λ'/d
λ'/λ = w'/w
or λ' =λ (w'/w) = 6000 × 0.7 = 4200 Å
From the above observation we conclude that, the wavelength of light used would be 4200 Å.
____________________________________________________________________________________________
A beam of light consisting of two wavelengths 6500A°and 5200A° is used to obtain interference fringes in a Young’s double slit experiment
(a) Find the distance of the third fringe on the screen from the central maximum for the wavelength 6500A°.
(b) What is the least distance from the central maximum where the bright fringes due to both wavelength coincide?
(c) The distance between the slits is 2mm and the distance between the plane of the slits and screen is 120cm. What is the fringe width for l = 6500A°?
(a) The width of the fringe Dλ/d
Then distance of the third fringe
3w = 3Dλ/d = [(3) (120) (6500) (10^{-8})]/[0.2] = 0.117 cm
Thus, the distance of the third fringe on the screen from the central maximum for the wavelength 6500A° would be 0.117 cm.
(b) Let m^{th} and n^{th} bright fringe of the wavelength coincide. Now position m^{th} bright fringe is
y_{m} = mλ_{1} (D/d)
and
y_{n}_{ }= nλ_{2} (D/d)
Or, m/n = λ_{1}/λ_{2 }= 5200/6500 = 4/5
(c) Fringe width, w = λD/d
= [6500 × 10^{-10} ×1.2] / [2× 10^{-13}] = 3900× 10^{-7} m = 0.039 cm
From the above observation we conclude that, the fringe width for l = 6500A° would be 0.039 cm.
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