Question

Where would an object be placed in a medium of refractive index ${\mu }_1$ , so that its real image is formed at equidistant from sphere (of radius R and refractive index ${\mu }_2$) which is also placed in the medium of refractive index ${\mu }_1$ as shown in figure?

• A. $\left(\frac{{\mu }_2-{\mu }_1}{{\mu }_2+{\mu }_1}\right)R$
• B. $\left(\frac{{\mu }_2}{{\mu }_2-{\mu }_1}\right)R$
• C. $\left(\frac{{\mu }_1}{{\mu }_2-{\mu }_1}\right)R$
• D. $\left(\frac{{\mu }_2+{\mu }_1}{{\mu }_2-{\mu }_1}\right)R$

Answer 1

Answer: (C)

$\left(\frac{{\mu }_1}{{\mu }_2-{\mu }_1}\right)R$

Since the image is formed equidistant from the sphere as the distance of the object.

In order to form the image at the same distance, the ray should emerge out of the sphere at the same angle as its incident angle. To obtain this, the ray should be parallel to the axis of the sphere as shown in the figure.

Now, applying the formula for refraction at the spherical surface.

$\frac{{\mu }_2}{v}-\frac{{\mu }_1}{u}=\frac{{\mu }_2-{\mu }_1}{R}$

Since the ray becomes parallel, it will form the image at infinite.

$\frac{{\mu }_2}{+\infty }-\frac{{\mu }_1}{-x}=\frac{{\mu }_2-{\mu }_1}{R}$

So, the distance of the object can be obtained as:

$x=\frac{{\mu }_{1}R}{{\mu }_2-{\mu }_1}$

Option $\left(C\right)$ is correct.