Science Part i Solutions Solutions for Class 10 Science Chapter 6 - Refraction Of Light

Answer:

a. Refractive index depends on the wavelength of light.
b. The change in direction of light rays while going from one medium to another is called refraction.

Answer:

a.

Let $\mu$ be the refractive index of the glass slab.  Then, according to Snell's law,
$\frac{\sin i}{\sin r_1}=\mu$    .....(i)
and 
$\frac{\sin r_2}{\sin e}=\frac{1}{\mu }$   .....(ii)
But, r₁ = r₂   .....(iii)
Putting (iii) in (i), we have
$\frac{\sin i}{\sin r_2}=\mu$   .....(iv)
Multiplying (i) and (iv), we have
$$\begin{gathered} \frac{\sin i}{\sin e}=1 \\ \mathrm{or} \\ i=e \end{gathered}$$

b.

After rainfall, the tiny droplets of water present in the atmosphere act as a prism for the rays coming from the Sun. Thus, the sunlight after striking the surface of the droplets gets refracted and dispersed into its seven components as shown in the figure (figure showing just two components). After this, the light rays are subjected to total internal reflection. Then the rays are again refracted when they come out of the water droplet. Hence, rainbow formation is the combined effect of the refraction, dispersion, and total internal reflection of light.

Answer:

A. The correct reason for the twinkling of stars is changing refractive index of the atmospheric gases.

Light coming from the stars undergoes refraction on entering the Earth’s atmosphere. This refraction continues until it reaches the Earth’s surface. This happens because of temperature variation of atmospheric air. Hence, the atmospheric air has changing refractive index at various altitudes. In this case, starlight continuously travels from a rarer medium to a denser medium. Hence, it continuously bends towards the normal. The refractive index of air medium gradually increases with a decrease in altitude. The continuous bending of starlight towards the normal results in a slight rise of the apparent position of the star.

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The rays of light from the Sun travel in straight line until they reach the Earth's atmosphere. The rays of light from the Sun enter obliquely in the Earth's atmosphere. The light rays coming from the Sun bend because of refraction, and this bending increases further because of the further increase in the refractive index of the successive layers. This causes the light rays to bend and we see the Sun early. Similarly, at sunset, the apparent position of the Sun is visible to us and not the actual position because of the same bending of light rays effect. Thus, due to refraction we see the Sun rise about two minutes before it is actually there and during sunset, we see it for around two minutes more, even though it has already moved from that position.

C. ${}_a\mu _g=\frac{{\mu }_g}{{\mu }_a}=\frac{3}{2}$
where,
 $$\begin{gathered} {}_a\mu _g=\mathrm{Refractive} \mathrm{index} \mathrm{of} \mathrm{glass} \mathrm{w}.\mathrm{r}.\mathrm{t}. \mathrm{air} \\ {\mu }_{\mathrm{g}}=\mathrm{Refractive} \mathrm{index} \mathrm{of} \mathrm{glass} \\ {\mu }_{\mathrm{a}}=\mathrm{Refractive} \mathrm{index} \mathrm{of} \mathrm{air} \end{gathered}$$
Thus, 
${}_g\mu _a=\frac{{\mu }_a}{{\mu }_g}=\frac{2}{3}$
where, ${}_g\mu _a=\mathrm{Refractive} \mathrm{index} \mathrm{of} \mathrm{air} \mathrm{w}.\mathrm{r}.\mathrm{t}. \mathrm{glass}$
Hence, the correct answer is option d.

Answer:

a.
 $$\begin{gathered} \mathrm{Absolute} \mathrm{refractive} \mathrm{index} \mathrm{of} \mathrm{a} \mathrm{medium}, \mu =\frac{\mathrm{Speed} \mathrm{of} \mathrm{light} \mathrm{in} \mathrm{air}}{\mathrm{Speed} \mathrm{of} \mathrm{light} \mathrm{in} \mathrm{the} \mathrm{medium}} \\ \Rightarrow \mu =\frac{3\times {10}^8}{1.5\times {10}^8}=2 \end{gathered}$$

b.
 $$\begin{gathered} {}_{\mathrm{a}}\mu _{\mathrm{g}}=\frac{3}{2} \mathrm{and} {}_{\mathrm{a}}\mu _{\mathrm{w}}=\frac{4}{3} \\ {}_{\mathrm{w}}\mu _{\mathrm{g}}=? \end{gathered}$$
where, 
$$\begin{gathered} {}_{\mathrm{w}}\mu _{\mathrm{g}}=\mathrm{Refractive} \mathrm{index} \mathrm{of} \mathrm{glass} \mathrm{w}.\mathrm{r}.\mathrm{t}. \mathrm{water} \\ {}_{\mathrm{a}}\mathrm{\mu }_{\mathrm{g}}=\mathrm{Absolute} \mathrm{refractive} \mathrm{index} \mathrm{of} \mathrm{glass} \\ {}_{\mathrm{a}}\mathrm{\mu }_{\mathrm{w}}=\mathrm{Absolute} \mathrm{refractive} \mathrm{index} \mathrm{of} \mathrm{water} \end{gathered}$$
Now,
$$\begin{gathered} {}_{\mathrm{a}}\mu _{\mathrm{g}}=\frac{3}{2} \mathrm{and} {}_{\mathrm{a}}\mu _{\mathrm{w}}=\frac{4}{3} \\ {}_{\mathrm{w}}\mu _{\mathrm{g}}=\frac{{}_{\mathrm{a}}\mu _{\mathrm{g}}}{{}_{\mathrm{a}}\mu _{\mathrm{w}}}=\frac{3}{2}\times \frac{3}{4}=\frac{9}{8} \end{gathered}$$