Physics Q102 – AP-EAMCET 2026

A convex lens forms a real image and a virtual image of same size of an object placed separately at distances

u_{1}

and

u_{2}

respectively from the lens. Then the focal length of the lens is

Clue 1 of 3: Lens Magnification

Understand the concept of magnification for lenses, specifically how it relates to image size and whether the image is real or virtual.

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Step 1: Define Magnification for Real and Virtual Images✦ Active

For a convex lens, a real image is inverted, so its magnification $m_{1}$ is negative. A virtual image is erect, so its magnification $m_{2}$ is positive. The problem states the images are of the 'same size', which means the magnitude of magnification is equal: $| m_{1} | = | m_{2} |$ . Using the lens formula $\frac{1}{v} - \frac{1}{u} = \frac{1}{f}$ and magnification $m = \frac{v}{u}$ , we can express magnification in terms of object distance $u$ and focal length $f$ . From the lens formula, we can write $v = \frac{u f}{u + f}$ . Substituting this into the magnification formula, we get $m = \frac{f}{u + f}$ . When using the Cartesian sign convention where object distance $u$ is negative (e.g., $- u_{m a g}$ where $u_{m a g}$ is the magnitude), the formula becomes $m = \frac{f}{- u_{m a g} + f} = \frac{f}{f - u_{m a g}}$ .

Step 2: Apply Magnification for the Real Image○ Expand

For the real image, the object is placed at a distance $u_{1}$ . According to the sign convention, the object position is $- u_{1}$ . The magnification for a real image formed by a convex lens is negative: $m_{1} = \frac{f}{f - u_{1}}$ . Since $u_{1} > f$ for a real image, $f - u_{1}$ is negative, making $m_{1}$ negative, which is consistent. The magnitude of magnification is $| m_{1} | = | \frac{f}{f - u_{1}} | = \frac{f}{u_{1} - f}$ .

💡 Teacher's Secret Hint

Remember that for a real image formed by a convex lens, the object must be placed beyond the focal point, i.e., $u_{1} > f$ .

Step 3: Apply Magnification for the Virtual Image○ Expand

For the virtual image, the object is placed at a distance $u_{2}$ . According to the sign convention, the object position is $- u_{2}$ . The magnification for a virtual image formed by a convex lens is positive: $m_{2} = \frac{f}{f - u_{2}}$ . Since $u_{2} < f$ for a virtual image, $f - u_{2}$ is positive, making $m_{2}$ positive, which is consistent. The magnitude of magnification is $| m_{2} | = \frac{f}{f - u_{2}}$ .

💡 Teacher's Secret Hint

For a virtual image formed by a convex lens, the object must be placed between the optical center and the focal point, i.e., $u_{2} < f$ .

Step 4: Equate Magnitudes and Solve for Focal Length○ Expand

Given that the images are of the same size, we equate the magnitudes of magnification:

| m_{1} | = | m_{2} |

\frac{f}{u_{1} - f} = \frac{f}{f - u_{2}}

Since $f \neq 0$ , we can cancel $f$ from both sides:

\frac{1}{u_{1} - f} = \frac{1}{f - u_{2}}

Cross-multiply:

f - u_{2} = u_{1} - f

Rearrange the terms to solve for $f$ :

2 f = u_{1} + u_{2}

f = \frac{u_{1} + u_{2}}{2}

💡 Teacher's Secret Hint

Ensure careful algebraic manipulation to avoid sign errors when rearranging the terms.

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Physics Question 102 – AP-EAMCET 2026

Clue 1 of 3: Lens Magnification

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