JEE Physics 2026 Question with Solution

Given: A convex lens has power $P = +5 \, \text{D}$. The recorded pairs of object distance and image distance are $(25,96)$, $(30,62)$, $(35,37)$, $(45,35)$ and $(50,32)$.

Find: Which reading is inconsistent with the lens formula.

First, find the focal length from power:

$$P = \frac{1}{f}$$

So,

$$f = \frac{1}{P} = \frac{1}{5} \, \text{m} = 0.2 \, \text{m} = 20 \, \text{cm}$$

Now use the lens formula:

$$\frac{1}{f} = \frac{1}{v} + \frac{1}{u}$$

Hence for a correct reading,

$$\frac{1}{20} = \frac{1}{v} + \frac{1}{u}$$

Check each person one by one.

For $P_1$:

$$\frac{1}{25} + \frac{1}{96} \approx \frac{1}{20}$$

So this reading is correct.

For $P_2$:

$$\frac{1}{30} + \frac{1}{62} \approx \frac{1}{20}$$

So this reading is correct.

For $P_3$:

$$\frac{1}{35} + \frac{1}{37} \neq \frac{1}{20}$$

So this reading is incorrect.

For $P_4$ and $P_5$, the given values satisfy the lens formula approximately, so these readings are correct.

Therefore, only $P_3$ has recorded incorrect readings. The correct option is D.

Given: The focal length is $20 \, \text{cm}$.

Find: Whether each measured pair matches the image distance expected from the lens formula.

Using

$$\frac{1}{f} = \frac{1}{v} + \frac{1}{u}$$

we get

$$\frac{1}{v} = \frac{1}{20} - \frac{1}{u}$$

For $u = 25 \, \text{cm}$:

$$\frac{1}{v} = \frac{1}{20} - \frac{1}{25} = \frac{1}{100}$$

so

$$v = 100 \, \text{cm}$$

The recorded value is $96 \, \text{cm}$, which is close enough experimentally.

For $u = 30 \, \text{cm}$:

$$\frac{1}{v} = \frac{1}{20} - \frac{1}{30} = \frac{1}{60}$$

so

$$v = 60 \, \text{cm}$$

The recorded value is $62 \, \text{cm}$, which is close enough.

For $u = 35 \, \text{cm}$:

$$\frac{1}{v} = \frac{1}{20} - \frac{1}{35} = \frac{3}{140}$$

so

$$v = \frac{140}{3} \, \text{cm} \approx 46.7 \, \text{cm}$$

But the recorded value is $37 \, \text{cm}$, which is not consistent.

For $u = 45 \, \text{cm}$:

$$\frac{1}{v} = \frac{1}{20} - \frac{1}{45} = \frac{1}{36}$$

so

$$v = 36 \, \text{cm}$$

The recorded value is $35 \, \text{cm}$, which is close.

For $u = 50 \, \text{cm}$:

$$\frac{1}{v} = \frac{1}{20} - \frac{1}{50} = \frac{3}{100}$$

so

$$v = \frac{100}{3} \, \text{cm} \approx 33.3 \, \text{cm}$$

The recorded value is $32 \, \text{cm}$, which is also close.

Hence the only incorrect reading is that of $P_3$. The correct option is D.