AP EAMCET · PHYSICS · Ray Optics

A lens forms real and virtual images of an object when the object is at $u_{1}$ and $u_{2}$ distances respectively. If the size of the virtual image is double that of the real image, then the focal length of the lens is (Take the magnification of the real image as $' m'$ )

A $(\frac{u_{1} + u_{2}}{2}) m$
B $(\frac{u_{1} - u_{2}}{3}) 2 m$
C $(\frac{u_{1} - u_{2}}{2}) 3 m$
D $(\frac{u_{1} + u_{2}}{3}) 2 m$

Verified Solution

Answer & Solution

Correct Answer

(B) $(\frac{u_{1} - u_{2}}{3}) 2 m$

Step-by-step Solution

Detailed explanation

The formula of focal length for a convex lens is, 1f=1v-1u, and formula for magnification, m=vu. For real image, u1-ve, v1+ve⇒1f=1v-1u⇒1f=1v1+1u1⇒u1f=u1v1+1⇒u1f=1m+1 ...Eq1 Similarly, for virtual image case, we can write…

Apply the relevant identity from the chapter and substitute the values established in the previous step, keeping the original constraint in view.

Use the simplified expression to isolate the unknown, then plug the result back into the question setup to confirm the working is internally consistent.

Cross-check against a boundary or limiting case. Both the dimensional balance and the qualitative behaviour at the extreme should agree with the candidate answer.

Combine the steps above to identify the correct option. The remaining choices each fail at least one of the consistency, dimensional, or boundary checks.

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A lens forms real and virtual images of an object when the object is at u1 and u2 distances respectively. If the size of the virtual image is double that of the real image, then the focal length of the lens is (Take the magnification of the real image as 'm')

Step-by-step Solution

A lens forms real and virtual images of an object when the object is at $u_{1}$ and $u_{2}$ distances respectively. If the size of the virtual image is double that of the real image, then the focal length of the lens is (Take the magnification of the real image as $' m'$ )