Practicing Success

Target Exam

CUET

Subject

Physics

Chapter

Ray Optics

Question:

CMP: The following figure shows a simple version of a zoom

The converging lens has focal length $f_{1’}$ and the diverging lens has focal length $f_2 = –|f_2|$. The two lenses are separated by a variable distance d that is always less than $f_{1’}$ also the magnitude of the focal length of the diverging lens satisfies the inequality $|f_2| > (f_1 –d)$. If the rays that emerge from the diverging lens and reach the final image point are extended backward to the left to the diverging lens, they will eventually expand to the original radius $r_0$ at the same point Q. To determine the effective focal length of the combination lens, consider a bundle of parallel rays of radius $r_0$ entering the converging lens.

Find the effective focal length?

Options:

$\frac{f_1|f_2|}{|f_2|-f_1+d}$

$\frac{f_1f_2}{f_1-f_2-d}$

$\frac{f_1}{f_2-f_1+d}$

$\frac{f_2}{f_1+f_2+d}$

Correct Answer:

$\frac{f_1|f_2|}{|f_2|-f_1+d}$

Explanation:

From the similar triangles in the sketch

$\frac{r_0}{r}=\frac{r'_0}{s'_2}$ so $f=\frac{f_1(f_2)}{|f_2|-f_1+d}$