Question

An object is placed between the concave lens and concave mirror. Object is at a distance d/3 from mirror and distance d from lend as shown in the figure. If final image formed by the given combination councides with the object, then focal length of lens is?

pls give detailed solution ​

Answer 1

Answer:

Focal length of the concave lens is given as -11d/5

Explanation:

first reflection by the mirror the image will form at

$\frac{1}{d_i} + \frac{1}{d_o} = \frac{1}{f}$

$\frac{1}{d_i} - \frac{3}{d} = -\frac{1}{d}$

$d_i = \frac{d}{2}$

now this image will act as an object for lens

So for lens again we have

$\frac{1}{d_i} + \frac{1}{d_o} = \frac{1}{f}$

so we have

$d_o = d + d/3 + d/2 = 11d/6$

$d_i = d</p><p>so plug in all values in above equation</p><p>[tex]\frac{1}{d} - \frac{6}{11d} = \frac{1}{f}$

$f = -\frac{11d}{5}$

#Learn

Topic : Combination of mirror and lens

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