A diverging lens with magnitude of focal length 25 cm is placed at a distance of 15 cm from a converging lens of magnitude of focal length 20 cm. A beam of parallel light falls on the diverging lens. The final image formed is:

We have two thin lenses placed on the same principal axis. The first lens met by the light is a diverging (concave) lens whose magnitude of focal length is 25 cm. By the sign convention for lenses, the focal length of a diverging lens is taken as negative, so we write

$$f_1=-25\ \text{cm}.$$

The second lens, 15 cm to the right of the first, is a converging (convex) lens with magnitude of focal length 20 cm. A converging lens has a positive focal length, so

$$f_2=+20\ \text{cm}.$$

The separation of the two lenses is

$$d=15\ \text{cm}.$$

A parallel beam of light is incident from the left on the diverging lens. For parallel rays falling on any thin lens, the image is formed at the focal point. Therefore, after refraction through the diverging lens the rays appear to come from its focus on the same side as the incident light. Hence the first image $$I_1$$ is virtual, 25 cm in front of the diverging lens:

$$u_1=\infty,\qquad v_1=f_1=-25\ \text{cm}.$$

This virtual image now acts as the object for the converging lens. We must find the object distance $$u_2$$ for the second lens. Because the two lenses are 15 cm apart and $$I_1$$ is 25 cm to the left of the first lens, the distance of $$I_1$$ from the second lens is

$$u_2=-(25+15)\ \text{cm}=-40\ \text{cm}.$$

(The negative sign arises because, by convention, an object situated to the left of a lens has a negative object distance.)

Now we apply the thin-lens formula to the converging lens. The formula is

$$\frac{1}{v}-\frac{1}{u}=\frac{1}{f}.$$

Substituting $$u=u_2=-40\ \text{cm}$$ and $$f=f_2=+20\ \text{cm}$$, we obtain

$$\frac{1}{v_2}-\frac{1}{(-40)}=\frac{1}{20}.$$

Simplifying,

$$\frac{1}{v_2}+\frac{1}{40}=\frac{1}{20}.$$

Bringing the fractions to a common denominator gives

$$\frac{1}{v_2}=\frac{1}{20}-\frac{1}{40} =\frac{2-1}{40} =\frac{1}{40}.$$

Hence

$$v_2=+40\ \text{cm}.$$

The positive value of $$v_2$$ means that the final image formed by the converging lens lies on the side opposite to the incident light, i.e. to the right of the converging lens. Therefore the image is real and is located 40 cm from the convergent lens.

Hence, the correct answer is Option B.