Physics, asked by sukhinder1279, 1 year ago

State huygens principle in wave optics . how did hygen explain the absence of backwave. used this principle to draw the refracted wave front for a plane wave incident from densor to rarer medium hence obtain snell's law of refraction

Answers

Answered by jomolmary8
3
Every point on the primary wave front act as the source of secondary wavelet, which travels forward with the same velocity of the original flame. This is known as Huygen's principle of secondary wavelets.
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Answered by vikrambrainly
0

Answer:

The frequency of incident, reflected and refracted light are all same as it only depends on the source of light.

Explanation:

Each point on the primary wave acts as a source of secondary wavelets. The new wavefront at any instant is the envelope of secondary wavelets at that instant.

Step 1: Time taken for light to go from Q to $Q^{\prime}$

\mathrm{t}=\frac{\mathrm{QK}}{\mathrm{c}}+\frac{\mathrm{Q} \mathrm{K}}{\mathrm{v}}

In right angled triangle $\triangle \mathrm{AQK}, \angle \mathrm{QAK}=\mathrm{i}$

$\therefore \mathrm{QK}=\mathrm{AK} \sin$ i...........(ii)

In right angled triangle $\triangle \mathrm{P}^{\prime} \mathrm{Q}^{\prime} \mathrm{K}, \angle \mathrm{Q}^{\prime} \mathrm{P}^{\prime} \mathrm{K}=\mathrm{r}$ and $\mathrm{Q}^{\prime} \mathrm{K}=\mathrm{KP} \mathrm{Pin}^{\prime} \mathrm{r}$.

(iii)

Step 2: Using (i),(ii) and (iii) we get

$$\begin{aligned}& \mathrm{t}=\frac{\mathrm{AK} \sin \mathrm{i}}{\mathrm{c}}+\frac{K P^{\prime} \sin \mathrm{r}}{v} \\& \mathrm{t}=\frac{\mathrm{AK} \sin \mathrm{i}}{\mathrm{c}}+\frac{\left(\mathrm{AP}^{\prime}-\mathrm{AK}\right) \sin \mathrm{r}}{v}\left(\because \mathrm{KP}^{\prime}=\mathrm{AP}^{\prime}-\mathrm{AK}\right) \\& \mathrm{t}=\frac{\mathrm{AP}^{\prime} \sin \mathrm{r}}{v}+\mathrm{AK}\left(\frac{\sin \mathrm{i}}{\mathrm{c}}-\frac{\sin \mathrm{r}}{v}\right)\end{aligned}$$

Step 3: The incident rays will fall at the same time at corresponding points of the refracted wavefront if it is independent of AK

i.e. $\frac{\sin i}{c}-\frac{\sin r}{v}=0$

$\frac{\sin \mathrm{i}}{\sin \mathrm{r}}=\frac{\mathrm{c}}{\mathrm{v}} \Rightarrow \mathrm{n}=\frac{\sin \mathrm{i}}{\sin \mathrm{r}}$ which is the Snell's Law for refraction of light.

(b)(i) The frequency of incident, reflected and refracted light are all same as it only depends on the source of light.

(ii)There is no reduction in energy since frequency remains constant.

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