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**Hint:**To solve this question, you need to recall how the wavelength varies as we go from one end to the other of VIBGYOR, and how the refractive index changes with wavelength as it increases or decreases. Also know that if the refractive index decreases, the critical angle increases and vice-versa.

**Complete step by step answer:**

As explained in the hint section of the solution, we will first talk about the change in wavelength as we go from one end to the other of VIBGYOR, and how this change in wavelength affects the refractive index which in turn affects the critical angle for the light rays.

So, as we all know, if we go from one end to the other of VIBGYOR, the wavelength is not the same throughout and varies as we move. If we go from Violet to Red, the wavelength increases, hence, we can observe that the Red light rays have the highest wavelength while the Violet light rays have the least wavelength. Using this information, we can observe that the wavelength of Yellow colour is lower than the wavelength of the Red colour.

Now, we need to check how the wavelength affects the refractive index. From Bragg's law, it has already been derived that wavelength of light affects the refractive index. Not just this, we have also developed the fact that as the wavelength increases, the refractive index decreases and as the wavelength decreases, the refractive index increases. Now, we already know that the wavelength of Red colour is higher than the wavelength of the Yellow colour, hence, we can confidently say that the refractive index of Red colour is lower than the refractive index of the Yellow colour.

Let us now use Snell's law to find out how refractive index affects the critical angle.

Snell’s law can be mathematically given as:

$\dfrac{{\sin i}}{{\sin r}} = \dfrac{{{n_2}}}{{{n_1}}}$

Where, $i$ is the angle of incidence,

$r$ is the angle of refraction,

${n_2}$ is the refractive index of the secondary medium, i.e. into which the light rays are going and,

${n_1}$ is the primary medium, i.e. the medium from which the light rays are coming.

For critical angle:

$

i = {i_c} \\

r = {90^ \circ } \\

$

And as the secondary medium in the question is given to be air, we can assume that:

${n_2} = 1$

Substituting the above-mentioned values, we get:

$\sin {i_c} = \dfrac{1}{{{n_1}}}$

Now, for the Yellow colour, we can write:

$\sin {i_y} = \dfrac{1}{{{n_y}}}$

Similarly, for Red colour, we can write:

$\sin {i_r} = \dfrac{1}{{{n_r}}}$

We know that ${n_r} < {n_y}$

So, $\dfrac{1}{{{n_r}}} > \dfrac{1}{{{n_y}}}$

From this, we can say that:

$\sin {i_r} > \sin {i_y}$

Which implies that:

${i_r} > {i_y}$

Or, that critical angle for the colour Red would be greater than the critical angle for the colour Yellow.

**Hence, the correct answer is option (A) since the critical angle for Red colour is more than ${45^ \circ }$.**

**Note:**Many students believe that since both Red and Yellow colour are constituents of the White light, they both would behave exactly the same and thus, they tick the option which says “same as ${45^ \circ }$“ as the answer, which causes them to lose marks. Even though both the colours are constituents of the White light, they have different wavelengths, and thus, behave differently.

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