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Natalija [7]
3 years ago
13

Problem One: A beam of red light (656 nm) enters from air into the side of a glass and then into water. wavelength, c. and speed

in both the glass and the water. Find the a. the frequencies, b. the
Physics
1 answer:
Ivanshal [37]3 years ago
4 0

Answer:

Part a)

f_w = f_g = 4.57 \times 10^{14} Hz

Part b)

\lambda_w = 492 nm

\lambda_g = 437.3 nm

Part c)

v_w = 2.25 \times 10^8 m/s

v_g = 2.0 \times 10^8 m/s

Explanation:

Part a)

frequency of light will not change with change in medium but it will depend on the source only

so here frequency of light will remain same in both water and glass and it will be same as that in air

f = \frac{v}{\lambda}

f = \frac{3 \times 10^8}{656 \times 10^{-9}}

f = 4.57 \times 10^{14} Hz

Part b)

As we know that the refractive index of water is given as

\mu_w = 4/3

so the wavelength in the water medium is given as

\lambda_w = \frac{\lambda}{\mu_w}

\lambda_w = \frac{656 nm}{4/3}

\lambda_w = 492 nm

Similarly the refractive index of glass is given as

\mu_w = 3/2

so the wavelength in the glass medium is given as

\lambda_g = \frac{\lambda}{\mu_g}

\lambda_g = \frac{656 nm}{3/2}

\lambda_g = 437.3 nm

Part c)

Speed of the wave in water is given as

v_w = \frac{c}{\mu_w}

v_w = \frac{3 \times 10^8}{4/3}

v_w = 2.25 \times 10^8 m/s

Speed of the wave in glass is given as

v_g = \frac{c}{\mu_g}

v_g = \frac{3 \times 10^8}{3/2}

v_g = 2 \times 10^8 m/s

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