In an experiment a convex lens of focal length 15 cm is placed coaxially on an optical bench in front of a convex mirror at a distance of 5 cm from it. It is found that an object and its image coincide, if the object is placed at a distance of 20 cm from the lens. The focal length of the convex mirror is-

The object is first kept $$20\;{\rm cm}$$ to the left of the thin convex lens and the focal length of the lens is given as $$f_L=+15\;{\rm cm}.$$

For a thin lens we have the lens formula

$$\frac1v-\frac1u=\frac1f,$$

where $$u$$ is the object-distance from the lens, $$v$$ is the image-distance from the lens and $$f$$ is the focal length of the lens. According to the usual Cartesian sign convention, distances measured to the left of the lens are negative and those measured to the right are positive.

Here the object lies to the left of the lens, so $$u=-20\;{\rm cm}.$$ Substituting in the lens formula we get

$$\frac1v-\frac1{(-20)}=\frac1{+15}$$

$$\Longrightarrow\;\frac1v+\frac1{20}=\frac1{15}$$

$$\Longrightarrow\;\frac1v=\frac1{15}-\frac1{20}$$

$$\Longrightarrow\;\frac1v=\frac{4}{60}-\frac{3}{60}=\frac1{60}$$

$$\Longrightarrow\;v=+60\;{\rm cm}.$$

Thus the lens produces its first image $$I_1$$ at a point $$60\;{\rm cm}$$ to the right of the lens.

The convex mirror is placed $$5\;{\rm cm}$$ to the right of the lens. Therefore the distance of the image $$I_1$$ from the mirror is

$$MI_1 = 60\;{\rm cm}-5\;{\rm cm}=55\;{\rm cm}.$$

Now observe the important experimental fact given in the problem: after reflection from the mirror and a second passage through the lens the rays form an image that coincides with the original object itself. For the rays to retrace their complete path (so that the final image sits exactly on the object) the rays must hit the mirror normally and return along the same lines. A spherical mirror sends rays back along their own paths only when the source of the rays is located at its centre of curvature. Hence the point $$I_1$$ must be the centre of curvature $$C$$ of the convex mirror.

Consequently

$$MC=MI_1=55\;{\rm cm}.$$

The radius of curvature of the mirror is therefore

$$R=55\;{\rm cm}.$$

For any spherical mirror the relation between its focal length $$f_M$$ and its radius of curvature $$R$$ is

$$f_M=\frac{R}{2}.$$

Substituting $$R=55\;{\rm cm}$$ gives

$$f_M=\frac{55\;{\rm cm}}{2}=27.5\;{\rm cm}.$$

Because the mirror is convex, the focal length is positive, so the numerical value we have obtained is already the required answer.

Hence, the correct answer is Option D.