Question

Light of wavelength 4000A° in medium 1 is incident on a plane boundary between media 1 and 2. As it enters from medium 1 into 2, its speed increases by 25% and its frequency in medium 2

is found to be 5 × 10^14 Hz. The absolute refractive index of medium 1 is​

Answer 1

Answer:

very very sorry I don't know

Answer 2

Concept

Frequency is that the number of occurrences of a repeating event per unit of your time. it's also occasionally remarked as temporal frequency to emphasise the contrast to spatial frequency, and ordinary frequency to stress the contrast to angular frequency.

Given

The wavelength of sunshine is $4000A^{\circ}$. The speed increases by $25\%$ from medium $1$ into $2$ and its frequency is $5\times 10^{14}HZ$ in medium $2.$

Find

We have to seek out absolutely the index of refraction of the medium $1.$

Solution

Frequency, being a characteristic of the source of sunshine doesn't change with the change of the medium. However, wavelength and speed both change because the medium changes. Here the frequency of sunshine within the medium $2$ is $5\times 10^{14}Hz$.

Hence, the frequency of given light in medium $1$ is also $5\times 10^{14}Hz$.

Further, wavelength in medium $1$ being $4000A^{\circ}$, speed of sunshine in medium $1.$

$\begin{aligned}v_{1}&=\left(5 \times 10^{14}\right)\left(4000\times 10^{-10}\right)m/sec\\ &=2\times 10^{8} m/sec\end{aligned}$

The absolute refractive index of medium $1$ is

$\begin{aligned}\mu_{1}&=\frac{c}{v_{1}}\\ &=\frac{3\times 10^{8}}{2\times 10^8}\\ &=1.5\end{aligned}$

Hence, the absolute refractive index of medium $1$ is $1.5.$

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