QJEEM21Ph49c7: JEE Main 2021 Physics - Class 12 JEEM21Ph49c7 Solution: Region I and II are separated by a spherical surface of radius…

As we know that, the equation of refraction at spherical surface is :

$$\frac{{\mu }_{II}}{v}-\frac{{\mu }_I}{u}=\frac{{\mu }_{II}-{\mu }_I}{R}$$

where,

$$\mu$$_II = refractive index of region II = 1.4

$$\mu$$_I = refractive index of region I = 1.25

R = radius of curvature = $$-$$25 cm

u = object distance = $$-$$40 cm

and v = image distance.

$$\Rightarrow \frac{1.4}{v}-\frac{1.25}{-40}=\frac{1.4-1.25}{-25}$$

$$\Rightarrow \frac{1.4}{v}=\frac{-0.15}{25}-\frac{1.25}{40}$$ $$\Rightarrow$$ v = $$-$$ 37.58 cm

where, negative sign indicate real image.