1. Given data

• Convex lens focal length: $f_L = 15\,\text{cm}$
• Distance between lens and convex mirror: $d = 5\,\text{cm}$
• Object is placed $20\,\text{cm}$ in front of the lens
• Object and final image coincide

We need the focal length of the convex mirror.

2. First image formed by the lens

Using the lens formula:

Using Cartesian sign convention for the lens:

So,

Hence,

So, if the mirror were absent, the lens would form an image $60\,\text{cm}$ to the right of the lens.

3. Object for the convex mirror

The mirror is only $5\,\text{cm}$ to the right of the lens, while the lens alone would form the image at $60\,\text{cm}$ to the right of the lens.

Therefore, relative to the mirror, this point is:

behind the mirror.

So for the mirror, the object is virtual and lies $55\,\text{cm}$ behind it. Thus,

Let the image formed by the mirror be at distance $v_m$ from the mirror.

4. Condition for final image to coincide with the original object

The original object is $20\,\text{cm}$ to the left of the lens. For the final image to coincide with the object after reflection and second refraction through the lens, the light returning from the mirror must form, for the lens on second pass, an image at $20\,\text{cm}$ to the left of the lens.

That means for the second pass through the lens, the rays must be incident on the lens as if coming from a point on its right side such that the lens forms image at $20\,\text{cm}$ left.

Let this point be the object for the lens on return, at distance $u_2$ to the right of the lens. Then for the lens,

Now for second pass:

• object is on right side of lens, so $u_2 = +x$
• final image coincides with original object, i.e. $v_2 = -20\,\text{cm}$

Thus,

This gives a sign inconsistency if interpreted directly; instead use the symmetry/reversibility idea: for the final image to retrace back to the object point, after reflection the mirror must form an image at the first image position corresponding to the object point for reverse travel through the lens.

So let us find where an object placed at $20\,\text{cm}$ left of the lens would need a source on the right so that after passing through lens it goes to that object point. By reversibility, this is the same as the first image position for object at $20\,\text{cm}$ left, i.e. $60\,\text{cm}$ right. Thus the mirror must send rays back as if diverging from the same point $60\,\text{cm}$ right of lens.

But that would not produce coincidence at the object after reflection unless the mirror forms image at a point whose conjugate through the lens is the object. The standard way is: after reflection, the mirror image acts as object for the lens at distance

to the right of mirror? Let us compute carefully.

The final image coincides with the original object, which is $20\,\text{cm}$ left of the lens. Therefore, for the second pass through the lens, the object for the lens must be at the point whose image through lens is at $20\,\text{cm}$ left.

Using lens formula with desired image $v_2=-20\,\text{cm}$:

This means the object for the second pass must be virtual, lying $\frac{60}{7}$ cm to the left of lens, which is not compatible with mirror side geometry. So use ray reversibility in a simpler way:

If object and final image coincide, the combined lens-mirror-lens system behaves like a ray retracing system. Therefore the image formed by the lens alone at $60\,\text{cm}$ right of lens must be reflected by the mirror back into itself. Thus the mirror must form image of the virtual object at $55\,\text{cm}$ behind mirror at the same point.

For a mirror, if object and image coincide, the object lies at center of curvature:

Hence,

Since it is a convex mirror, the focal length magnitude is

5. Checking options

• A: $27.5\,\text{cm}$ ✅
• B: $20.0\,\text{cm}$ ❌
• C: $25.0\,\text{cm}$ ❌
• D: $30.5\,\text{cm}$ ❌

So the correct option is A.

6. Comparison with stored answer

Stored correct answer: A

My derived answer: A

They agree.