Question

A thin equiconvex glass lens $({\mu }_g=1.5)$ is being placed on the top of a vessel of height $h=20$ cm as shown in fig. A luminous point source is being placed at the bottom of vessel on the principal axis of the lens. When the air is on both the side of the lens the image of luminous source is formed at a distance of 20 cm from the lens out side the vessel. When the air inside the vessel is being replaced by a liquid of refractive index ${\mu }_1$, the image of the same source is being formed at a distance 30 cm from the lens outside the vessel. If ${\mu }_1$ is approximately $\frac{x}{100}$. Find $x$.

Answer 1

1. When air is on both sides of the lens,

$\frac{1}{20}-\frac{1}{-20}=\frac{1}{f}$

Hence, $\frac{1}{f}=\frac{1}{10}=({\mu }_{lens}-1)(\frac{1}{R}-\frac{1}{-R})$

Hence, $R=10cm$

2. When liquid is inside the vessel.

For the refraction from bottom side of the lens,

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

hence $\frac{1.5}{v}=\frac{3-3{\mu }_1}{20}$

Now this v acts as an object for the refraction from top surface of the lens.

Thus
$\frac{1}{v_1}-\frac{1.5}{v}=\frac{1-1.5}{-10}$

Final image is formed 30cm from lens,

Thus $v_1=30cm$

Hence, ${\mu }_1=1.11$