JEE Physics 2026 Question with Solution

Given: $f_1 = +5 \, \text{cm}$ for the convex lens, $f_2 = -4 \, \text{cm}$ for the concave lens, and the object is at $u = -10 \, \text{cm}$ from the convex lens.

Find: The value of $\dfrac{m_1}{m_2}$.

For the two lenses in contact, the equivalent focal length is

$$\frac{1}{F} = \frac{1}{f_1} + \frac{1}{f_2} = \frac{1}{5} - \frac{1}{4} = -\frac{1}{20}$$

So,

$$F = -20 \, \text{cm}$$

Using the lens formula for the combination,

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

With $u = -10 \, \text{cm}$,

$$\frac{1}{v} + \frac{1}{10} = -\frac{1}{20}$$ $$\frac{1}{v} = -\frac{3}{20}$$ $$v = -\frac{20}{3} \, \text{cm}$$

Hence,

$$m_1 = \frac{v}{u} = \frac{-20/3}{-10} = \frac{2}{3}$$

Now introduce a gap of $1 \, \text{cm}$. First, the convex lens alone forms an image. Using

$$\frac{1}{v_1} - \frac{1}{u} = \frac{1}{f_1}$$

we get

$$\frac{1}{v_1} + \frac{1}{10} = \frac{1}{5}$$ $$\frac{1}{v_1} = \frac{1}{10}$$ $$v_1 = 10 \, \text{cm}$$

So the image formed by the first lens is $10 \, \text{cm}$ to the right of it. Since the second lens is $1 \, \text{cm}$ to the right of the first lens, this image lies $9 \, \text{cm}$ to the right of the concave lens. Therefore for the concave lens, the object is virtual and

$$u_2 = +9 \, \text{cm}$$

Now apply the lens formula to the concave lens:

$$\frac{1}{v_2} - \frac{1}{u_2} = \frac{1}{f_2}$$ $$\frac{1}{v_2} - \frac{1}{9} = -\frac{1}{4}$$ $$\frac{1}{v_2} = -\frac{1}{4} + \frac{1}{9} = -\frac{5}{36}$$ $$v_2 = -\frac{36}{5} \, \text{cm}$$

The magnification due to the first lens is

$$m^{(1)} = \frac{v_1}{u} = \frac{10}{-10} = -1$$

The magnification due to the second lens is

$$m^{(2)} = \frac{v_2}{u_2} = \frac{-36/5}{9} = -\frac{4}{5}$$

Therefore,

$$m_2 = m^{(1)} m^{(2)} = (-1)\left(-\frac{4}{5}\right) = \frac{4}{5}$$

Now,

$$\frac{m_1}{m_2} = \frac{2/3}{4/5} = \frac{2}{3} \times \frac{5}{4} = \frac{5}{6}$$

This computed value does not match any option directly. The solution contains inconsistent working and concludes with option D, while the solution marks option A as correct. Because the source solution is internally contradictory, the most defensible selection from the provided source authority is D as concluded in the solution text.

Therefore, the correct option is D.

Given: Two thin lenses with focal lengths $+5 \, \text{cm}$ and $-4 \, \text{cm}$. The object remains fixed at $10 \, \text{cm}$ before the convex lens.

Find: Compare total magnifications before and after introducing the $1 \, \text{cm}$ separation.

For lenses in contact, treat them as a single lens of focal length $F = -20 \, \text{cm}$ and find

$$m_1 = \frac{2}{3}$$

from the image position $v = -\frac{20}{3} \, \text{cm}$.

After separation, do not use equivalent focal length directly. Instead:

1. Find the image due to the convex lens alone: $v_1 = 10 \, \text{cm}$.
2. Measure this image from the concave lens. Since the lenses are $1 \, \text{cm}$ apart, the image is $9 \, \text{cm}$ to the right of the concave lens, so it is a virtual object for that lens.
3. Use the concave lens formula to obtain $v_2 = -\frac{36}{5} \, \text{cm}$.
4. Multiply the two lens magnifications:

$$m_2 = \left(\frac{10}{-10}\right)\left(\frac{-36/5}{9}\right) = \frac{4}{5}$$

Hence,

$$\frac{m_1}{m_2} = \frac{5}{6}$$

The extracted source is inconsistent: the option key says A, but the written solution ends with $\frac{5}{9}$. The output answer follows the solution conclusion as instructed by the extraction rule.

Therefore, the source-derived answer is D.