HSC Science (Electronics) 12th Board ExamMaharashtra State Board
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# Solution for Determine the change in wavelength of light during its passage from air to glass. If the refractive index of glass with respect to air is 1.5 and the frequency of light is 3.5 x 1014 Hz - HSC Science (Electronics) 12th Board Exam - Physics

ConceptReflection and Refraction of Plane Wave at a Plane Surface Using Wave Fronts

#### Question

Determine the change in wavelength of light during its passage from air to glass. If the refractive index of glass with respect to air is 1.5 and the frequency of light is 3.5 x 1014 Hz, find the wave number of light in glass.

[Velocity of light in air c = 3 x 108 m/s]

#### Solution

Given: μg = 1.5, n=4 × 1014Hz, c = 3 × 108m/s

The wavelength of light incident on glass from air is

lambda=c/n=(3 xx 10^8)/(3.5 xx 10^14)=8.571 xx 10^-7 m=8571 xx 10^-10 m=8571Å

Now, the velocity of light in glass is given from its refractive index as

mu_g = c/(v_g)

We also know that velocity is product of frequency and wavelength.

∴mu=c/(v_g)=(nlambda_a)/(nlambda_g)=lambda_a/(lambdag)

∴lambda_g=lambda_a/mu_g=8571 /1.5=5714Å

Therefore, the difference in wavelength is

lambda_a-lambda_g=8571 - 5714 = 2857Å

The wave number is the reciprocal of the wavelength

Therefore, the wave number in glass is

barlambda_g=1/lambda_g

∴barlambda_g=1/(5.714 xx 10^-7)=1.75 xx 10^6m^-1

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Solution for question: Determine the change in wavelength of light during its passage from air to glass. If the refractive index of glass with respect to air is 1.5 and the frequency of light is 3.5 x 1014 Hz concept: Reflection and Refraction of Plane Wave at a Plane Surface Using Wave Fronts. For the courses HSC Science (Electronics), HSC Science (General) , HSC Science (Computer Science)
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