Fresnel's distance (ZF) is the distance for which the ray optics is a good approximation. It is given by the relation. ZF = a²/λ Where, Aperture width, a = 4 mm = 4 x 1o-3 m Wavelength of light, λ = 400 nm = 400 x 10-9 m ZF = (4 x 1o-3 m)²/(400 x 10-9) = 40m Therefore, the distance for which the rayRead more
Fresnel’s distance (ZF) is the distance for which the ray optics is a good approximation. It is given by the relation.
ZF = a²/λ
Where, Aperture width, a = 4 mm = 4 x 1o-3 m
Wavelength of light, λ = 400 nm = 400 x 10-9 m
ZF = (4 x 1o-3 m)²/(400 x 10-9) = 40m
Therefore, the distance for which the ray optics is a good approximation is 40 m.
Wavelength of incident light, λ = 5000 Aº = 5000 x 10⁻10 m Speed of light, c = 3x 10s m Frequency of incident light is given by the relation, ν =c/λ =3x 108 /5000 x 10⁻10 = 6 x 10¹⁴ Hz The wavelength and frequency of incident light is the same as that of reflected ray. Hence, the wavelength of reflRead more
Wavelength of incident light, λ = 5000 Aº = 5000 x 10⁻10 m Speed of light, c = 3x 10s m Frequency of incident light is given by the relation, ν =c/λ =3x 108 /5000 x 10⁻10 = 6 x 10¹⁴ Hz The wavelength and frequency of incident light is the same as that of reflected ray. Hence, the wavelength of reflected light is 5000 A and its frequency is 6 x 1014 Hz. When reflected ray is normal to incident ray, the sum of the angle of incidence, i and angle of reflection, r is 90°. According to the law of reflection, the angle of incidence is always equal to the angle of reflection. Hence, we can write the sum as: <i + <r = 90 <i + <i = 90
2<i = 90
Therefore, <i = 90/2 = 45º
Therefore, the angle of incidence for the given condition is 45°.
Refractive index of glass, μ = 1.5 Brewster angle = 0 Brewster angle is related to refractive index as: tan 0 = μ => 0 = tan-1 (1.5) = 56.31° Therefore, the Brewster angle for air to glass transition is 56.31°.
Refractive index of glass, μ = 1.5
Brewster angle = 0
Brewster angle is related to refractive index as: tan 0 = μ
=> 0 = tan-1 (1.5) = 56.31°
Therefore, the Brewster angle for air to glass transition is 56.31°.
Distance of the screen from the slits, D = 1 m Wavelength of light used, λ₁= 600 nm Angular width of the fringe in air, 0₁= 0.2° Angular width of the fringe in water = 02 Refractive index of water, μ=4/3 Refractive index is related to angular width as: μ=0₁/02 => 02= 3/4 0₁ = 3/4 x 0.2 = 0.15 ThRead more
Distance of the screen from the slits, D = 1 m
Wavelength of light used, λ₁= 600 nm
Angular width of the fringe in air, 0₁= 0.2°
Angular width of the fringe in water = 02
Refractive index of water, μ=4/3
Refractive index is related to angular width as:
μ=0₁/02
=> 02= 3/4 0₁ = 3/4 x 0.2 = 0.15
Therefore, the angular width of the fringe in water will reduce to 0.15°.
Wavelength of the light beam, λ1= 650 nm Wavelength of another light beam, λ2 = 520 nm Distance of the slits from the screen = D Distance between the two slits = d Ans (a). Distance of the nth bright fringe on the screen from the central maximum is given by the relation, x = n λ1 (D/d) For third briRead more
Wavelength of the light beam, λ1= 650 nm
Wavelength of another light beam, λ2 = 520 nm
Distance of the slits from the screen = D
Distance between the two slits = d
Ans (a).
Distance of the nth bright fringe on the screen from the central maximum is given by the relation,
x = n λ1 (D/d)
For third bright fringe, n = 3
Therefore , x = 3 x 650 D/d = 1950 D/d nm
Ans (b).
Let the nth bright fringe due to wavelength λ2 and (n – l)th bright fringe due to wavelength λ1 coincide on the screen. We can equate the conditions for bright fringes as:
nλ2 = (n -1) λ1
520 n = 650n – 650
=>650 = 130 n
Therefore ,n = 5
Hence, the least distance from the central maximum can be obtained by the relation:
x = n λ2 D/d
= 5×520 D/d= 2600 D/d nm
Note: The value of d and D are not given in the question.
Let I1and I2 be the intensity of the two lightwaves. Their resultant intensities can be obtained as: I' = I1 + I2 + 2√I1l2 cos 0 Where, 0 = Phase difference between the two waves For monochromatic light waves, I1 = I2 Therefore , I' = I1 + I1+ 2 √I1I1 cos 0 = 2I1+2I1 cos0 Phase difference = 2π/λ Read more
Let I1and I2 be the intensity of the two lightwaves. Their resultant intensities can be obtained as: I’ = I1 + I2 + 2√I1l2 cos 0
Where, 0 = Phase difference between the two waves For monochromatic light waves,
I1 = I2
Therefore , I’ = I1 + I1+ 2 √I1I1 cos 0
= 2I1+2I1 cos0
Phase difference = 2π/λ x path difference
Since path difference = λ,
Phase difference, 0 = 2π
Therefore ,I’ =2I1+2I1 =4 I1
Given I’ = K
Therefore ,I1= K/4 ——-Eq-1
When path difference = λ/3, the phase difference, 0 =2π/3
Distance between the slits, d = 0.28 mm = 0.28 x 10-3 m Distance between the slits and the screen, D = 1.4 m Distance between the central fringe and the fourth (n = 4) fringe, u = 1.2 cm = 1.2 x 10-2 m In case of a constructive interference, we have the relation for the distance between the two frinRead more
Distance between the slits, d = 0.28 mm = 0.28 x 10-3 m
Distance between the slits and the screen, D = 1.4 m
Distance between the central fringe and the fourth (n = 4) fringe, u = 1.2 cm = 1.2 x 10-2 m
In case of a constructive interference, we have the relation for the distance between the two fringes as: u = n λ D/d
Where, n = Order of fringes = 4 λ = Wavelength of light used.
Ans (a). Refractive index of glass, μ = 1.5 Speed of light, c = 3 x 108 m/s Speed of light in glass is given by the relation, v =c/μ = (3x108)/1.5 = 2 x 108 m/s. Hence, the speed of light in glass is 2 x 108 m/s. Ans (b). The speed of light in glass is not independent of the colour of light. The refRead more
Ans (a).
Refractive index of glass, μ = 1.5
Speed of light, c = 3 x 108 m/s
Speed of light in glass is given by the relation,
v =c/μ
= (3×108)/1.5 = 2 x 108 m/s.
Hence, the speed of light in glass is 2 x 108 m/s.
Ans (b).
The speed of light in glass is not independent of the colour of light.
The refractive index of a violet component of white light is greater than the refractive index of a red component. Hence, the speed of violet light is less than the speed of red light in glass. Hence, violet light travels slower than red light in a glass prism.
Wavelength of incident monochromatic light, λ = 589 nm = 589 x 10-9 m Speed of light in air, c = 3 x 108 m/s Refractive index of water, μ = 1.33 Ans (a). The ray will reflect back in the same medium as that of incident ray. Hence, the wavelength, speed, and frequency of the reflected ray will be theRead more
Wavelength of incident monochromatic light, λ = 589 nm = 589 x 10-9 m
Speed of light in air, c = 3 x 108 m/s
Refractive index of water, μ = 1.33
Ans (a).
The ray will reflect back in the same medium as that of incident ray. Hence, the wavelength, speed, and frequency of the reflected ray will be the same as that of the incident ray.
Frequency of light is given by the relation,
ν = c/λ = 3 x 108 /(589 x 10-9 )
=5.09 x 10¹4Hz
Hence, the speed, frequency, and wavelength of the reflected light are 3 x 108 m/s, 5.09 x 1014 Hz, and 589 nm respectively.
Ans (b).
Frequency of light does not depend on the property of the medium in which it is travelling. Hence, the frequency of the refracted ray in water will be equal to the frequency of the incident or reflected light in air.
Refracted frequency, v= 5.09 x 1014 Hz
Speed of light in water is related to the refractive index of water as:
ν = c/μ
ν = 3 x 108/1.33 =2.26 x 108m/s
Wavelength of light in water is given by the relation,
λ =ν /v = (2.26 x 108) /(5.09 x 1014 )
= 444.007 x 10⁻9 m = 444.01 nm
Hence, the speed, frequency, and wavelength of refracted light are 2.26 x 108 m/s, 444.01 nm, and 5.09 x 1014 Hz respectively.
Focal length of the convex lens, f₁ = 30 cm The liquid acts as a mirror. Focal length of the liquid = f2 Focal length of the system (convex lens + liquid), f = 45 cm For a pair of optical systems placed in contact, the equivalent focal length is given as: 1/f =1/f ₁+ 1/f₂ 1/f₂ =1/f -1/f₁ => 1/45Read more
Focal length of the convex lens, f₁ = 30 cm
The liquid acts as a mirror. Focal length of the liquid = f2
Focal length of the system (convex lens + liquid), f = 45 cm
For a pair of optical systems placed in contact, the equivalent focal length is given as:
1/f =1/f ₁+ 1/f₂
1/f₂ =1/f -1/f₁
=> 1/45 -1/30 = -1/90
Therefore ,f₂ = -90 cm
Let the refractive index of the lens be μ₁ and the radius of curvature of one surface be R.
Hence, the radius of curvature of the other surface is -R.
R can be obtained using the relation:1/f₁ = (μ₁— 1) )1/R + 1/(-R)
=>1/30 = (1.5-1) (2/R)
=> R = 30/(0.5 x 2) = 30 cm Let μ2 be the refractive index of the liquid.
Radius of curvature of the liquid on the side of the plane mirror = ∝
Radius of curvature of the liquid on the side of the lens, R = -30 cm.
The value of μ2 can be calculated using the relation:
1/f2 = (μ2-1) [1/(-R) -1/∝
-1/90 = (μ2-1) [1/(+30) -0]
μ2-1 = 1/3
Therefore ,μ2 = 4/3 = 1.33
Hence, the refractive index of the liquid is 1.33.
Estimate the distance for which ray optics is good approximation for an aperture of 4 mm and wavelength 400 nm.
Fresnel's distance (ZF) is the distance for which the ray optics is a good approximation. It is given by the relation. ZF = a²/λ Where, Aperture width, a = 4 mm = 4 x 1o-3 m Wavelength of light, λ = 400 nm = 400 x 10-9 m ZF = (4 x 1o-3 m)²/(400 x 10-9) = 40m Therefore, the distance for which the rayRead more
Fresnel’s distance (ZF) is the distance for which the ray optics is a good approximation. It is given by the relation.
ZF = a²/λ
Where, Aperture width, a = 4 mm = 4 x 1o-3 m
Wavelength of light, λ = 400 nm = 400 x 10-9 m
ZF = (4 x 1o-3 m)²/(400 x 10-9) = 40m
Therefore, the distance for which the ray optics is a good approximation is 40 m.
See lessLight of wavelength 5000 Aº falls on a plane reflecting surface. What are the wavelength and frequency of the reflected light? For what angle of incidence is the reflected ray normal to the incident ray?
Wavelength of incident light, λ = 5000 Aº = 5000 x 10⁻10 m Speed of light, c = 3x 10s m Frequency of incident light is given by the relation, ν =c/λ =3x 108 /5000 x 10⁻10 = 6 x 10¹⁴ Hz The wavelength and frequency of incident light is the same as that of reflected ray. Hence, the wavelength of reflRead more
Wavelength of incident light, λ = 5000 Aº = 5000 x 10⁻10 m
Speed of light, c = 3x 10s m
Frequency of incident light is given by the relation,
ν =c/λ =3x 108 /5000 x 10⁻10 = 6 x 10¹⁴ Hz
The wavelength and frequency of incident light is the same as that of reflected ray. Hence, the wavelength of reflected light is 5000 A and its frequency is 6 x 1014 Hz. When reflected ray is normal to incident ray, the sum of the angle of incidence, i and angle of reflection, r is 90°.
According to the law of reflection, the angle of incidence is always equal to the angle of reflection. Hence, we can write the sum as:
<i + <r = 90
<i + <i = 90
2<i = 90
Therefore, <i = 90/2 = 45º
Therefore, the angle of incidence for the given condition is 45°.
See lessWhat is the Brewster angle for air to glass transition? (Refractive index of glass = 1.5.)
Refractive index of glass, μ = 1.5 Brewster angle = 0 Brewster angle is related to refractive index as: tan 0 = μ => 0 = tan-1 (1.5) = 56.31° Therefore, the Brewster angle for air to glass transition is 56.31°.
Refractive index of glass, μ = 1.5
Brewster angle = 0
Brewster angle is related to refractive index as: tan 0 = μ
=> 0 = tan-1 (1.5) = 56.31°
Therefore, the Brewster angle for air to glass transition is 56.31°.
See lessIn a double-slit experiment the angular width of a fringe is found to be 0.2° on a screen placed 1 m away. The wavelength of light used is 600 nm. What will be the angular width of the fringe if the entire experimental apparatus is immersed in water? Take refractive index of water to be 4/3.
Distance of the screen from the slits, D = 1 m Wavelength of light used, λ₁= 600 nm Angular width of the fringe in air, 0₁= 0.2° Angular width of the fringe in water = 02 Refractive index of water, μ=4/3 Refractive index is related to angular width as: μ=0₁/02 => 02= 3/4 0₁ = 3/4 x 0.2 = 0.15 ThRead more
Distance of the screen from the slits, D = 1 m
Wavelength of light used, λ₁= 600 nm
Angular width of the fringe in air, 0₁= 0.2°
Angular width of the fringe in water = 02
Refractive index of water, μ=4/3
Refractive index is related to angular width as:
μ=0₁/02
=> 02= 3/4 0₁ = 3/4 x 0.2 = 0.15
Therefore, the angular width of the fringe in water will reduce to 0.15°.
See lessA beam of light consisting of two wavelengths, 650 nm and 520 nm, is used to obtain interference fringes in a Young’s double-slit experiment. (a) Find the distance of the third bright fringe on the screen from the central maximum for wavelength 650 nm. (b) What is the least distance from the central maximum where the bright fringes due to both the wavelengths coincide?
Wavelength of the light beam, λ1= 650 nm Wavelength of another light beam, λ2 = 520 nm Distance of the slits from the screen = D Distance between the two slits = d Ans (a). Distance of the nth bright fringe on the screen from the central maximum is given by the relation, x = n λ1 (D/d) For third briRead more
Wavelength of the light beam, λ1= 650 nm
Wavelength of another light beam, λ2 = 520 nm
Distance of the slits from the screen = D
Distance between the two slits = d
Ans (a).
Distance of the nth bright fringe on the screen from the central maximum is given by the relation,
x = n λ1 (D/d)
For third bright fringe, n = 3
Therefore , x = 3 x 650 D/d = 1950 D/d nm
Ans (b).
Let the nth bright fringe due to wavelength λ2 and (n – l)th bright fringe due to wavelength λ1 coincide on the screen. We can equate the conditions for bright fringes as:
nλ2 = (n -1) λ1
520 n = 650n – 650
=>650 = 130 n
Therefore ,n = 5
Hence, the least distance from the central maximum can be obtained by the relation:
x = n λ2 D/d
= 5×520 D/d= 2600 D/d nm
Note: The value of d and D are not given in the question.
See lessIn Young’s double-slit experiment using monochromatic light of wavelength λ, the intensity of light at a point on the screen where path difference is λ, is K units. What is the intensity of light at a point where path difference is λ/3?
Let I1and I2 be the intensity of the two lightwaves. Their resultant intensities can be obtained as: I' = I1 + I2 + 2√I1l2 cos 0 Where, 0 = Phase difference between the two waves For monochromatic light waves, I1 = I2 Therefore , I' = I1 + I1+ 2 √I1I1 cos 0 = 2I1+2I1 cos0 Phase difference = 2π/λ Read more
Let I1and I2 be the intensity of the two lightwaves. Their resultant intensities can be obtained as: I’ = I1 + I2 + 2√I1l2 cos 0
Where, 0 = Phase difference between the two waves For monochromatic light waves,
I1 = I2
Therefore , I’ = I1 + I1+ 2 √I1I1 cos 0
= 2I1+2I1 cos0
Phase difference = 2π/λ x path difference
Since path difference = λ,
Phase difference, 0 = 2π
Therefore ,I’ =2I1+2I1 =4 I1
Given I’ = K
Therefore ,I1= K/4 ——-Eq-1
When path difference = λ/3, the phase difference, 0 =2π/3
Hence, resultant intensity, I’R = I1 + I2 + 2√I1I2 cos 2π/3
= 2I1 + 2I1 (-1/2) = I1
Using equation (1), we can write:
IR = I1=K/4
Hence, the intensity of light at a point where the path difference is λ/3 is K/4 units.
See lessIn a Young’s double-slit experiment, the slits are separated by 0.28 mm and the screen is placed 1.4 m away. The distance between the central bright fringe and the fourth bright fringe is measured to be 1.2 cm. Determine the wavelength of light used in the experiment.
Distance between the slits, d = 0.28 mm = 0.28 x 10-3 m Distance between the slits and the screen, D = 1.4 m Distance between the central fringe and the fourth (n = 4) fringe, u = 1.2 cm = 1.2 x 10-2 m In case of a constructive interference, we have the relation for the distance between the two frinRead more
Distance between the slits, d = 0.28 mm = 0.28 x 10-3 m
Distance between the slits and the screen, D = 1.4 m
Distance between the central fringe and the fourth (n = 4) fringe, u = 1.2 cm = 1.2 x 10-2 m
In case of a constructive interference, we have the relation for the distance between the two fringes as: u = n λ D/d
Where, n = Order of fringes = 4 λ = Wavelength of light used.
Therefore ,λ = ud/(nD)
= (1.2 x 10-2 x 0.28 x 10-3)/(4 x 1.4)
4×1.4
= 6 x 10–7 = 600 nm
Hence, the wavelength of the light is 600 nm.
See less(a) The refractive index of glass is 1.5. What is the speed of light in glass? Speed of light in vacuum is 3.0 x 10⁸ m s⁻¹) (b) Is the speed of light in glass independent of the colour of light? If not, which of the two colours red and violet travels slower in a glass prism?
Ans (a). Refractive index of glass, μ = 1.5 Speed of light, c = 3 x 108 m/s Speed of light in glass is given by the relation, v =c/μ = (3x108)/1.5 = 2 x 108 m/s. Hence, the speed of light in glass is 2 x 108 m/s. Ans (b). The speed of light in glass is not independent of the colour of light. The refRead more
Ans (a).
Refractive index of glass, μ = 1.5
Speed of light, c = 3 x 108 m/s
Speed of light in glass is given by the relation,
v =c/μ
= (3×108)/1.5 = 2 x 108 m/s.
Hence, the speed of light in glass is 2 x 108 m/s.
Ans (b).
The speed of light in glass is not independent of the colour of light.
The refractive index of a violet component of white light is greater than the refractive index of a red component. Hence, the speed of violet light is less than the speed of red light in glass. Hence, violet light travels slower than red light in a glass prism.
See lessMonochromatic light of wavelength 589 nm is incident from air on a water surface. What are the wavelength, frequency and speed of (a) reflected, and (b) refracted light? [Refractive index of water is 1.33.]
Wavelength of incident monochromatic light, λ = 589 nm = 589 x 10-9 m Speed of light in air, c = 3 x 108 m/s Refractive index of water, μ = 1.33 Ans (a). The ray will reflect back in the same medium as that of incident ray. Hence, the wavelength, speed, and frequency of the reflected ray will be theRead more
Wavelength of incident monochromatic light, λ = 589 nm = 589 x 10-9 m
Speed of light in air, c = 3 x 108 m/s
Refractive index of water, μ = 1.33
Ans (a).
The ray will reflect back in the same medium as that of incident ray. Hence, the wavelength, speed, and frequency of the reflected ray will be the same as that of the incident ray.
Frequency of light is given by the relation,
ν = c/λ = 3 x 108 /(589 x 10-9 )
=5.09 x 10¹4Hz
Hence, the speed, frequency, and wavelength of the reflected light are 3 x 108 m/s, 5.09 x 1014 Hz, and 589 nm respectively.
Ans (b).
Frequency of light does not depend on the property of the medium in which it is travelling. Hence, the frequency of the refracted ray in water will be equal to the frequency of the incident or reflected light in air.
Refracted frequency, v= 5.09 x 1014 Hz
Speed of light in water is related to the refractive index of water as:
ν = c/μ
ν = 3 x 108/1.33 =2.26 x 108m/s
Wavelength of light in water is given by the relation,
λ =ν /v = (2.26 x 108) /(5.09 x 1014 )
= 444.007 x 10⁻9 m = 444.01 nm
Hence, the speed, frequency, and wavelength of refracted light are 2.26 x 108 m/s, 444.01 nm, and 5.09 x 1014 Hz respectively.
See lessFigure shows an equiconvex lens (of refractive index 1.50) in contact with a liquid layer on top of a plane mirror. A small needle with its tip on the principal axis is moved along the axis until its inverted image is found at the position of the needle. The distance of the needle from the lens is measured to be 45.0 cm. The liquid is removed and the experiment is repeated. The new distance is measured to be 30.0 cm. What is the refractive index of the liquid?
Focal length of the convex lens, f₁ = 30 cm The liquid acts as a mirror. Focal length of the liquid = f2 Focal length of the system (convex lens + liquid), f = 45 cm For a pair of optical systems placed in contact, the equivalent focal length is given as: 1/f =1/f ₁+ 1/f₂ 1/f₂ =1/f -1/f₁ => 1/45Read more
Focal length of the convex lens, f₁ = 30 cm
The liquid acts as a mirror. Focal length of the liquid = f2
Focal length of the system (convex lens + liquid), f = 45 cm
For a pair of optical systems placed in contact, the equivalent focal length is given as:
1/f =1/f ₁+ 1/f₂
1/f₂ =1/f -1/f₁
=> 1/45 -1/30 = -1/90
Therefore ,f₂ = -90 cm
Let the refractive index of the lens be μ₁ and the radius of curvature of one surface be R.
Hence, the radius of curvature of the other surface is -R.
R can be obtained using the relation:1/f₁ = (μ₁— 1) )1/R + 1/(-R)
=>1/30 = (1.5-1) (2/R)
=> R = 30/(0.5 x 2) = 30 cm Let μ2 be the refractive index of the liquid.
Radius of curvature of the liquid on the side of the plane mirror = ∝
Radius of curvature of the liquid on the side of the lens, R = -30 cm.
The value of μ2 can be calculated using the relation:
1/f2 = (μ2-1) [1/(-R) -1/∝
-1/90 = (μ2-1) [1/(+30) -0]
μ2-1 = 1/3
Therefore ,μ2 = 4/3 = 1.33
Hence, the refractive index of the liquid is 1.33.
See less