Mastering the Two-Lens System
A Journey Through Geometrical Optics
Welcome to a fascinating exploration of geometrical optics! In this problem, we are tasked with analyzing a two-lens system. We are given four different combinations of lenses, and for each combination, we need to determine the exact position of the final image.
The setup is consistent across all four cases: an object is placed 20 cm to the left of the first lens, and the two lenses are separated by a fixed distance of 5 cm. To solve this, we will rely heavily on the fundamental lens formula:
By rearranging this equation, we can express the image distance v directly as:
This rearranged form will be our master tool, saving us precious time during calculations. Let's dive into the cases one by one.
Case I
Two Convex Lenses
In our first scenario, both lenses are convex. The first lens has a focal length f1=+10 cm. The object is placed to the left, so by standard sign convention, the object distance is u1=−20 cm.
Plugging these values into our master equation:
v1=−20+10(−20)(10)=−10−200=+20 cm
The positive sign indicates that the first image, I1, is formed 20 cm to the right of the first lens.
Now, here is the crucial conceptual leap: the image formed by the first lens acts as the object for the second lens. Since the lenses are 5 cm apart, and I1 is 20 cm to the right of the first lens, I1 lies 15 cm to the right of the second lens. Because the light rays are converging towards a point behind the second lens, I1 acts as a virtual object for the second lens. Therefore, u2=+15 cm.
The second lens is also convex, with f2=+15 cm. Applying our formula again:
v2=15+15(15)(15)=30225=+7.5 cm
The final image is formed 7.5 cm to the right of the second lens. This perfectly matches option (P).
Case II
A Convex and a Concave Lens
In the second case, the first lens is identical to the one in Case I (f1=+10 cm). Consequently, the first image is formed at the exact same location: v1=+20 cm.
The object distance for the second lens remains u2=+15 cm. However, the second lens is now a concave lens with a focal length f2=−10 cm. We must be incredibly careful with the negative sign here!
v2=15+(−10)(15)(−10)=5−150=−30 cm
The negative sign tells us that the final image is formed 30 cm to the left of the second lens. This matches option (R).
Case III
A Stronger Concave Lens
Once again, the first lens is unchanged, so v1=+20 cm and u2=+15 cm.
This time, the second lens is a weaker concave lens with f2=−20 cm. Let's see how this affects the final image:
v2=15+(−20)(15)(−20)=−5−300=+60 cm
The negatives cancel out beautifully, leaving us with a positive 60 cm. The final image is formed 60 cm to the right of the second lens, which corresponds to option (Q).
Case IV
Starting with a Concave Lens
In our final case, we flip the script. The first lens is now concave with f1=−20 cm. The object is still at u1=−20 cm.
v1=−20+(−20)(−20)(−20)=−40400=−10 cm
The first image is formed 10 cm to the left of the first lens.
Now, we must determine the object distance for the second lens. The image I1 is 10 cm to the left of the first lens, and the second lens is 5 cm to the right of the first lens. Therefore, the total distance from I1 to the second lens is 10+5=15 cm. Since I1 is to the left of the second lens, it acts as a real object, giving us u2=−15 cm.
The second lens is convex with f2=+10 cm.
v2=−15+10(−15)(10)=−5−150=+30 cm
The final image is formed 30 cm to the right of the second lens, matching option (T).
The Grand Conclusion
By meticulously applying the lens formula and strictly adhering to the sign convention, we have successfully mapped all four cases:
(I) maps to (P)
(II) maps to (R)
(III) maps to (Q)
(IV) maps to (T)
This problem is a masterclass in sequential optical systems. It teaches us that the output of one optical element seamlessly becomes the input for the next, provided we respect the geometry and the signs!
