Ray Optics and Optical Instruments
- Light travels through a glass plate of thickness t and refractive index µ. If c is the speed of light in vacuum, the time taken by light to travel this thickness of glass is
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Total thickness = t; Refrative index = µ
Speed of light in Glass plate = c μ 
∵ v = Speed of light in vacuum 
R.I. of medium Time taken = t = μt 
c 
c μ
[where, t = thickness of glass plate]Correct Option: D
Total thickness = t; Refrative index = µ
Speed of light in Glass plate = c μ ∵ v = Speed of light in vacuum R.I. of medium Time taken = t = μt c c μ
[where, t = thickness of glass plate]
- One face of a rectangular glass plate 6 cm thick is silvered. An object held 8 cm in front of the first face forms an image 12 cm behind the silvered face. The refractive index of the glass is
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Thickness of glass plate (t) = 6 cm;
Distance of the object (u) = 8 cm.
And distance of the image (v) = 12 cm.
Let x = Apparent position of the silvered surface in cm.
Since the image is formed due to reflection at the silvered face and by the property of mirror image
Distance of object from the mirror = Distance of image from the mirror
or, x + 8 = 12 + 6 – x ⇒ x = 5 cm. Therefore, refractive index of glass = Real depth = 6 = 1.2 Apparent depth 5 Correct Option: C
Thickness of glass plate (t) = 6 cm;
Distance of the object (u) = 8 cm.
And distance of the image (v) = 12 cm.
Let x = Apparent position of the silvered surface in cm.
Since the image is formed due to reflection at the silvered face and by the property of mirror image
Distance of object from the mirror = Distance of image from the mirror
or, x + 8 = 12 + 6 – x ⇒ x = 5 cm. Therefore, refractive index of glass = Real depth = 6 = 1.2 Apparent depth 5
- A beam of monochromatic light is refracted from vacuum into a medium of refractive index 1.5, the wavelength of refracted light will be
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From μ = c = n λv , λm = λv v n λm μ
Here, c = velocity of light in medium and v = velocity of light in vacuum;
µ = refractive index of the medium.
Hence, wavelength in medium (λm) < λa
(∵ µ > 1, given)
So, the required wavelength decreases.
ALTERNATIVELY,
c = vλ . On refraction, the frequency, do not change. When light is refracted from vacuum to a medium, the velocity, c decreases. Therefore , λ also decreases.
Correct Option: C
From μ = c = n λv , λm = λv v n λm μ
Here, c = velocity of light in medium and v = velocity of light in vacuum;
µ = refractive index of the medium.
Hence, wavelength in medium (λm) < λa
(∵ µ > 1, given)
So, the required wavelength decreases.
ALTERNATIVELY,
c = vλ . On refraction, the frequency, do not change. When light is refracted from vacuum to a medium, the velocity, c decreases. Therefore , λ also decreases.
- When a biconvex lens of glass having refractive index 1.47 is dipped in a liquid, it acts as a plane sheet of glass. This implies that the liquid must have refractive index.
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1 = μg - 1 1 - 1 f μm R1 R2 If µg = µm, then 1 = (1 - 1) 1 - 1 f R1 R2 ⇒ 1 = 0 f ⇒ f = 1 = ∞ 0
This implies that the liquid must have refractive index equal to glass.Correct Option: A
1 = μg - 1 1 - 1 f μm R1 R2 If µg = µm, then 1 = (1 - 1) 1 - 1 f R1 R2 ⇒ 1 = 0 f ⇒ f = 1 = ∞ 0
This implies that the liquid must have refractive index equal to glass.
- A plano convex lens fits exactly into a plano concave lens. Their plane surfaces are parallel to each other. If lenses are made of different materials of refractive indices µ1 and µ2 and R is the radius of curvature of the curved surface of the lenses, then the focal length of the combination is
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1 = 1 + 1 f f1 f2 = (μ1 - 1) 1 - 1 + (μ2 - 1) 1 - 1 ∞ -R ∞ R = (μ1 - 1) - (μ2 - 1) ⇒ 1 = μ1 - μ2 R R f R ⇒ f = R μ1 - μ2 Hence, focal length of the combination is R μ1 - μ2
Correct Option: B
1 = 1 + 1 f f1 f2 = (μ1 - 1) 1 - 1 + (μ2 - 1) 1 - 1 ∞ -R ∞ R = (μ1 - 1) - (μ2 - 1) ⇒ 1 = μ1 - μ2 R R f R ⇒ f = R μ1 - μ2 Hence, focal length of the combination is R μ1 - μ2
