(a)40 cm to left of 1st lens

(b)40 cm to right of 1st lens

(c)40 cm to left of second lens

(d)40 cm to right of second lens

The equivalent focal length is

$\large\frac{1}{F} =\frac{1}{f_1}+\frac{1}{f_2} -\frac{d}{f_1f_2}$

$ \qquad= \large\frac{1}{20}+\frac{1}{20} -\frac{60}{(20)^2}$

or $F=-20 \;cm$

It is divergent lens. It should be kept at a distance $D= \large\frac{dF}{F}$ behind second lens.

here $D= 60 \times \large\frac{(-20)}{20}$$=-60\;cm$

Thus equivalent divergent lens should be placed at a distance of 60 cm to right of second lens.

The final image is formed at focus of this divergent lens (ie) 20 cm to left of it .

Therefore It is 40 cm to right of second lens.

Hence d is the correct answer.

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