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(a) Sketch the refracted wavefront for the incident plane wavefront of light from a distant object passing through a convex lens.

(b) Using Huygens' principle, verify the laws of refraction when light from a denser medium is incident on a rarer medium.

(c) For yellow light of wavelength 590nm incident on a glass slab, the refractive index of glass is 1.5. Estimate the speed and wavelength of yellow light inside the glass slab.

Answers (1)

(a)

Refracted spherical wavefront emerges out which is converging towards the focus of a lens.

(b) According to Huygen's principle, each point of the wavefront is the source of the secondary disturbance and the wavelets emanating from these points spread out in all direction with the speed of the wave. These waves emanating from the wavefront are usually referred to as secondary wavelets and if we draw a common tangent to all these spheres, we obtain the new position of the wavefront at a later time.

Here V₂ > V₁ i.e., light ray travels from denser medium to rarer medium.

Let PP' be the surface separating medium 1 and medium 2.

Let V₁ and V₁ represent the speed of light in medium 1 and medium 2.

We assume that a plane wavefront AB is incident on the interface at an angle i.

Let $\tau$ be the time taken by the wavefront to travel the distance BC.

$\therefore BC = V_{1}\tau$

In order to determine the shape of the refracted wavefront, we draw a sphere radius from a point A in 2^ndmedium.

Let CE represents a tangent plane drawn from point C on the sphere.

$\therefore AE= V_{2}\tau$

CE represents refracted wavefront.

Now consider triangles ABC and AEC

$\sin i = \frac{BC}{AC}=\frac{V_{1}\tau}{AC}$

$\sin r = \frac{AE}{AC}=\frac{V_{2}\tau}{AC}$

$\angle i$ - incident angle

$\angle r$ - refracted angle

$\frac{\sin i}{\sin r}=\frac{\frac{V_{1}\tau}{AC}}{\frac{V_{2}\tau}{AC}}$

$n_{1}=\frac{C}{V_{1}}, n_{2}= \frac{C}{V_{2}}$

$\frac{\sin i}{\sin r}=\frac{\frac{c}{n_{1}}}{\frac{c}{n_{2}}}$

$\frac{\sin i}{\sin r}=\frac{n_{2}}{n_{1}}$

(c) Given

the wavelength of light in air is = 590nm

Refractive index of glass -=1.5

Speed of light in the glass, $V_{2}=\frac{c}{n}$

$V_{2}=\frac{3\times 10^{8}}{1.5}=2\times 10^{8}m/s$

$\frac{V_{1}}{\lambda _{1}}=\frac{V_{2}}{\lambda _{2}}$

$\frac{3\times 10^{8}}{590\times 10^{-9}}=\frac{2\times 10^{8}}{\lambda _{2}}$

$\therefore \lambda _{2}= 393\times 10^{-9}m =393 nm$

Posted by

Safeer PP

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