Question

The image of an object placed in the air formed by a convex refracting surface is at a distance of $10m$ behind the surface. The image is real and is at $2^{nd}/3$ of the distance of the object from the surface. The wavelength of light inside the surface is$2/3$ times the wavelength in air, The radius of the curved surface is $x/13m.$The value of ‘x’ is _______.

Answer 1

Step 1: Given data

The image of an object placed in the air formed by a convex refracting surface is at a distance $v=+10m$.

Let the distance of the object from the surface is $u$

The required diagram

The image is real and is at $2^{nd}/3$ of the distance of the object from the surface

$$\begin{gathered} v=\frac{2}{3}u \\ u=\frac{3}{2}v \\ u=\frac{3}{2}\times 10 \\ u=-15m \end{gathered}$$ from sign convention

The radius of the curved surface is $R=\frac{x}{13}m.$

The wavelength of light inside the surface is$2/3$ times the wavelength in air

Let ${\lambda }_m$ is the wavelength inside the surface.

${\lambda }_a$ is the wavelength in air.

${\lambda }_m=\frac{2}{3}{\lambda }_a$

Step 2: Formula used:

The refractive index is given by-

${\lambda }_m=\frac{{\lambda }_a}{\mu }\dots \dots \dots \dots \dots \dots \dots \dots \left(1\right)$

Where

$\mu$ is the refractive index

${\lambda }_m$ is the wavelength inside the surface.

${\lambda }_a$ is the wavelength in air.

From lens makers' formula

$\frac{\mu }{v}-\frac{1}{u}=\left(\frac{\mu -1}{R}\right)\dots \dots \dots \dots \dots \dots \dots \left(2\right)$

Where $R$ is radius of curvature

Step 3: Calculate the value of $x$.

First, find the value of the refractive index using the formula in equation (1)

$\begin{array}{rcl}{\lambda }_m& =& \frac{{\lambda }_a}{\mu }\\ \frac{2}{3}{\lambda }_a& =& \frac{{\lambda }_a}{\mu }\\ \mu & =& \frac{3}{2}\end{array}$ Substituting the given vaule ${\lambda }_m=\frac{2}{3}{\lambda }_a$

Now substitute the value in the lens makers' formula,

$\begin{array}{rcl}\frac{{\displaystyle \frac{3}{2}}}{10}-\frac{1}{-15}& =& \left(\frac{{\displaystyle \frac{3}{2}}-1}{R}\right)\\ \frac{{\displaystyle 3}}{20}+\frac{1}{15}& =& \left(\frac{{\displaystyle 1}}{2R}\right)\\ \frac{45+20}{300}& =& \left(\frac{{\displaystyle 1}}{2R}\right)\\ R& =& \frac{30}{13}\end{array}$

Compare the value that we got for the radius of curvature $R=\frac{30}{13}m$ with the given value of $R=\frac{x}{13}m.$

$$\begin{gathered} \frac{30}{13}=\frac{x}{13} \\ x=30 \end{gathered}$$

Therefore the value of $x=30$.