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.
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 less