Question

Ratio of refractive index of red light to blue light in air is:
A. less than unity.
B. equal to unity.
C. greater than unity.
D. less as well as greater than unity depending upon the experimental arrangement.

Answer 1

We will firstly write the equation representing the relation between wavelength of red and blue light and refractive index. After that we will analyse the relation and then will proceed further to obtain the result.

Formula used:
$\mu =\dfrac{c}{n\lambda }$

Complete answer:
We will assume that,
${{\lambda }_{r}}$be the wavelength of red light .
${{\lambda }_{b}}$ be the wavelength of blue light.
We know that wavelength of red is largest than wavelength of blue light
${{\lambda }_{r}}\succ {{\lambda }_{b}}$
Now from the formula
$\mu =\dfrac{c}{n\lambda }$
We know that wavelength is inversely proportional to refractive index.
So we obtain

& {{\mu }_{r}}\prec {{\mu }_{b}} \\\ & \dfrac{{{\mu }_{r}}}{{{\mu }_{b}}}\prec 1 \\\ \end{aligned}$$ So from above we obtain that ratio of refractive index of red light to blue light is less than 1 **Hence the correct option is A.** **Additional information:** One of the applications of this phenomenon is red light used as danger signals because red light has the higher wavelength and is scattered the least so red light can travel a larger distance irrespective of other lights in fog, rain and smoke to reach the eyes of the observer. Second application is the colour of the sky is blue. It can be explained as blue light has the shorter wavelength and is scattered most by the particles present in the air and hence the sky appears blue. **Note:** We know that the refractive index is defined as the ratio of the velocity of light travelling in vaccum to the ratio of light travelling in medium. And from this the relation between refractive index and wavelength can also be derived just by replacing velocity by wavelength. It can also be concluded that light with higher wavelength disperses more than light having lower wavelength.