Question

A concave spherical surface of radius of curvature $10cm$ separates two mediums $X$ and $Y$ of refractive indices $4/3$ and $3/2$ respectively. Centre of curvature of the surface lies in the medium $X$. An object is placed in medium $X$.

• A. Image is always real
• B. Image is real if the object distance is greater than $90cm$
• C. Image is always virtual
• D. Image is real if the object distance is less than $90cm$

Answer 1

Answer: (C)

It is given, that, a concave spherical surface of radius of curvature $10cm$ separates two medium of X and Y of refractive index $\frac{4}{3}$ and $\frac{3}{2}$ respectively.

Therefore, ${\mu }_X=\frac{4}{3}$ , ${\mu }_Y=\frac{3}{2}$

And, according to sign convention, $R=-10cm$.

Now, let us consider, the object is situated at a distance of $x$ in front of the concave spherical surface. So, object distance, $u=-x$ (using sign convention)

According to lens make's formula, we know,

$\frac{{\mu }_Y}{v}-\frac{{\mu }_X}{u}=\frac{{\mu }_Y-{\mu }_X}{R}$

Where, $v$ is the image distance.

$\therefore \frac{\frac{3}{2}}{v}-\frac{\frac{4}{3}}{-x}=\frac{\frac{3}{2}-\frac{4}{3}}{-10}=-\frac{1}{60}$

Or, $v=\frac{-1.5x}{0.017x+1.33}$

So, clearly, $v<0$

This implies that the image will always be a virtual one.

Therefore, the correct option is - (C) The image is always virtual.