Question

A concave refractive surface of a medium having refractive index $\mu$ produces a real image of an object (located outside the medium) irrespective of its location. Choose the correct option from the following
A. Always
B. May be if refractive index of surrounding medium is greater than $\mu$
C. May be if refractive index of surrounding medium is less than $\mu$
D. None of the above

Answer 1

For concave refractive surface, the radius of curvature, R and distance of object, u is negative. For refraction at a concave spherical surface, $\dfrac{{{\mu _2}}}{v} - \dfrac{{{\mu _1}}}{u} = \dfrac{{\left( {{\mu _2} - {\mu _1}} \right)}}{R}$ where ${\mu _1}$ and ${\mu _2}$ are the refractive index of the surrounding medium and the medium of refractive surface respectively. For concave refractive surfaces to produce a real image, the distance of the image, $v > 0$ .

Complete step by step solution:
In the question, it is given that a concave refractive surface of a medium having refractive index $\mu$ produces a real image of an object which is located outside the medium. Also, the location of the object is not specified.
We have the formula for any refractive surface $\dfrac{{{\mu _2}}}{v} - \dfrac{{{\mu _1}}}{u} = \dfrac{{{\mu _2} - {\mu _1}}}{R}$ where ${\mu _1}$ and ${\mu _2}$ are the refractive index of the surrounding medium and the medium of refractive surface respectively.
For concave refractive surface, the radius of curvature, R and distance of object, u is negative i.e. $R = - R$ and $u = - u$ .
Let the surrounding medium be air. So, ${\mu _1} = 1$ .
Also, ${\mu _2} = \mu$ as given in the question.
Substituting these value in the above equation, we get $\dfrac{\mu }{v} - \dfrac{1}{{ - u}} = \dfrac{{\mu - 1}}{{ - R}}$
Or, $\dfrac{\mu }{v} = \dfrac{{1 - \mu }}{R} - \dfrac{1}{u}$
Or, $\dfrac{\mu }{v} = \dfrac{{u\left( {1 - \mu } \right) - R}}{{uR}}$
Or, $v = \dfrac{{\mu uR}}{{u\left( {1 - \mu } \right) - R}}$
Now, as given in the question, the image produced is real. So, for a concave refractive surface to produce a real image, the distance of the image, $v > 0$ .
So, $\dfrac{{\mu uR}}{{u\left( {1 - \mu } \right) - R}} > 0$
For the above in equation to behold, $u\left( {1 - \mu } \right) - R > 0$
Or, $1 - \mu > \dfrac{R}{u}$
As, $\dfrac{R}{u} > 0$ , $1 - \mu > 0$
Or, $\mu < 1$ which shows that the refractive index of the surrounding medium is greater than $\mu$ .

$\therefore$The refractive index of the surrounding medium is greater than $\mu$. Hence, option B is correct.

Note:
Be careful while substituting the signs of radius of curvature (R), distance of the object (u) and distance of the image (v). Always remember the formula $\dfrac{{{\mu _2}}}{v} - \dfrac{{{\mu _1}}}{u} = \dfrac{{\left( {{\mu _2} - {\mu _1}} \right)}}{R}$ for refraction through curved surface.