Question

In the given arrangement, a lens of refractive index $1.5$ is placed having medium of refractive indices $n_1$ and $n_2$ on either sides as shown. The values of $n_1$ and $n_2$ respectively with respect to the lens are

• A. $1$ and $1$
• B. $1.5$ and $1.5$
• C. Less than $1.5$
• D. Greater than $1.5$ and less than $1.5$

Answer 1

The correct option is B $1.5$ and $1.5$

The nature of the lens will not change; shape, size, density etc will remain the same.

Now, the focal length will change.

The lens maker's formula is: $\frac{1}{f}=\frac{n_2-n_1}{n_1}(\frac{1}{R_1}-\frac{1}{R_2})$

where $f$ is the focal length, $n_2$ the refractive index of the lens, $n_1$ the refractive index of the medium and $R_1$ and $R_2$ the curvature radius of each lens's face.

In the question we can see there is no refraction in the light rays, it means the light rays will intersect at infinity. That means the focal lenth equals to $\infty$

If $f=\infty$ $\Rightarrow \frac{1}{f}=0$

Hence, $n_2=n_1$

And option $B$ is correct.