Reflection and Refraction of Light Waves Numerical Class-12 Nootan ISC Physics Solution Ch-19 Wave Nature of Light Huygens Principle. Step by step solutions of Kumar and Mittal Physics of Nageen Prakashan as council latest prescribe guideline for upcoming exam. Visit official Website CISCE for detail information about ISC Board Class-12 Physics.
Reflection and Refraction of Light Waves Numerical Class-12 Nootan ISC Physics Solution Ch-19 Wave Nature of Light Huygens Principle
| Board | ISC |
| Class | 12 |
| Subject | Physics |
| Book | Nootan |
| Chapter-19 | Wave Nature of Light Huygens Principle. |
| Topics | Numericals on Reflection and Refraction of Light Waves |
| Academic Session | 2025-2026 |
Numericals on Reflection and Refraction of Light Waves
Numerical Class-12 Nootan ISC Physics Solution Ch-19 Wave Nature of Light Huygens Principle
Que-1: What will be the speed of light in a medium of refractive index 1.5?
Ans- v = c/n
= 3 x 10^8/1.5
= 2 x 10^8 m/s
Que-2: Find the time taken by light to travel 1.0 m thickness of glass whose refractive index is 1.5.
Ans- v = c/n = 3x 10^8/1.5
= 2 x 10^8 m/s
t = d/v = 1/2 x 10^8
= 5 x 10^-9 s
Que-3: The index of refraction of glass for violet light is 1.665 and for red light 1.618. Compute the difference in the speed of light in glass for the two colours.
Ans- Vr – Vv = 3 x 10^8/1.618 – 3 x 10^8/1.665
= 5.2 x 10^6 m/s
Que-4: The refractive index of glass is 1.5 and that of water is 1.3. If the speed of light in water is 2.25 x 10^8 m/s, find the speed in glass.
Ans- Vg/Vw = nw/ng
=> Vg/2.25 x 10^8 = 1.3/1.5
=> Vg = 1.95 x 10^8 m/s
Que-5: The ratio of thicknesses of the strips of two transparent media A and B is 6: 4. If light takes the same time in passing through both of them, then what is the refractive index of B with respect to A?
Ans- t = dA/vA = dB/vB
=> AnB = vA/vB = dA/dB = 6/4 = 1.5
Que-6: Light of wavelength 4500 Å in vacuum enters into a glass block of refractive index 1.5. What is the wavelength of light in the glass block?
Ans- λ’ = λ/n
= 4500 / 1.5
= 3000 Å
Que-7: Green light of mercury has a wavelength of 5.5 x 10^-7 m in air. (i) What is its frequency in MHz and period in us? (ii) What is its wavelength in glass, if nglass = 1.5?
Ans- (i) f = c/λ
= 3 x 10^8 / 5.5 x 10^-7
= 5.45 x 10^14 Hz = 5.45 x 10^8 MHz
again T = 1/f
= 1/ 5.45 x 10^14
= 1.83 x 10^-15 s = 1.83 x 10^-9 μs
(ii) λ’ = λ/1.5
= 5.5 x 10^-7/1.5 = 3.67 x 10^-7 m
Que-8: The wavelength of monochromatic light in air is 6000 Å. The wavelength of this light in water is 4500 Å. Find out : (i) frequency of light in water, (ii) speed of light in water.
Ans- (i) frequency in water = frequency in air
= 3 x 10^8/6000 x 10^-10 = 5 x 10^14 Hz
again (ii) λ’/ λ = nw/na => va/vw = 6000/4000 = 3 x 10^8/vw
=> vw = 2.25 x 10^8 m/s
Que-9: Light waves of frequency 5.0 x 10^14 Hz pass through a liquid in which the wavelength of light is found to be 4.5 x 10^-7 m. Find : (i) wavelength of light in vacuum, (ii) refractive index of the liquid and (iii) velocity of light in the liquid.
Ans- n = λ/λ’ = c/v
(i) λ = c/f = 3 x 10^8/5 x 10^14
= 6 x 10^-7 m
(ii) n = 6 x 10^-7/4.5 x 10^-7
= 1.33
(iii) v = c/n = 3 x 10^8/1.33 = 2.25 x 10^8 m/s
Que-10: Light of wavelength 5000 Å is incident on a plane mirror. What are the wavelength and the frequency of the reflected light? For what angle of incidence is the reflected ray perpendicular to the incident ray?
Ans- After reflection wavelength and frequency remains same
therefore answer is 5000Å
again ∠i = ∠r
and ∠i + ∠r = 90°
=> ∠i = ∠r = 45°
Que-11: Yellow light of wavelength 600 nm is incident on a boundary separating air and glass. Find the speed, wavelength and frequency of: (i) reflected light and (ii) refracted light. The refractive index of glass for yellow light is 1.5.
Ans- (i) Reflection does not change frequency wavelength or speed
(ii) v = c/n = 3 x 10^8/1.5 = 2 x 10^8 m/s
frequency = 5.0 x 10^14 Hz (remains unchanged)
and wavelength,
λ’ = λ/n
= 600/1.5 = 400 nm
–: End of Reflection and Refraction of Light Waves Numerical Class-12 Nootan ISC Physics Solution :–
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