JEE Physics 2023 Question with Solution

Given: A thin plano-concave lens $L_1$ and a thin plano-convex lens $L_2$ have radius of curvature $R = 30 \, \text{cm}$, refractive index $\mu = 1.75$, and separation $40 \, \text{cm}$. The object is at infinity.

Find: The distance $x$ of the final image from the concave lens.

For the plano-concave lens,

$$\frac{1}{f_1} = (1 - \mu) \left(\frac{1}{R}\right)$$

Substituting the values,

$$\frac{1}{f_1} = (1 - 1.75) \cdot \frac{1}{30} = -\frac{0.75}{30}$$

Hence,

$$f_1 = -40 \, \text{cm}$$

For the plano-convex lens,

$$\frac{1}{f_2} = (\mu - 1) \left(\frac{1}{R}\right)$$

Substituting the values,

$$\frac{1}{f_2} = (1.75 - 1) \cdot \frac{1}{30} = \frac{0.75}{30}$$

Hence,

$$f_2 = 40 \, \text{cm}$$

Since the object is at infinity, the image formed by $L_1$ is at its focal point on the left, so this image is $40 \, \text{cm}$ to the left of $L_1$. As $L_2$ is $40 \, \text{cm}$ to the right of $L_1$, the object distance for $L_2$ is

$$u = -40 - 40 = -80 \, \text{cm}$$

Using the lens formula for $L_2$,

$$\frac{1}{v} - \frac{1}{u} = \frac{1}{f_2}$$

Substituting,

$$\frac{1}{v} - \frac{1}{-80} = \frac{1}{40}$$ $$\frac{1}{v} + \frac{1}{80} = \frac{1}{40}$$ $$\frac{1}{v} = \frac{1}{40} - \frac{1}{80} = \frac{1}{80}$$

Therefore,

$$v = 80 \, \text{cm}$$

So the final image is $80 \, \text{cm}$ to the right of $L_2$. From the concave lens $L_1$, the total distance is

$$x = 40 + 80 = 120 \, \text{cm}$$

Therefore, the image is formed at $120 \, \text{cm}$ from the concave lens.

Given: The first lens is plano-concave and the second lens is plano-convex. Both have $R = 30 \, \text{cm}$ and $\mu = 1.75$. The separation between the lenses is $40 \, \text{cm}$.

Find: The final image distance from the concave lens.

Step 1: Compute the focal lengths using the lens maker relation as used in the provided working.

$$\frac{1}{f_1} = (1 - \mu)\frac{1}{R} = (1 - 1.75)\frac{1}{30} = -\frac{0.75}{30}$$ $$f_1 = -40 \, \text{cm}$$$$\frac{1}{f_2} = (\mu - 1)\frac{1}{R} = (1.75 - 1)\frac{1}{30} = \frac{0.75}{30}$$ $$f_2 = 40 \, \text{cm}$$

Step 2: Since the object is at infinity, the concave lens forms a virtual image at its own focal point on the left side.

$$\text{Image due to } L_1 \text{ is at } 40 \, \text{cm} \text{ left of } L_1$$

Step 3: Measure the object distance for $L_2$. The second lens is $40 \, \text{cm}$ right of $L_1$, so the virtual object is $80 \, \text{cm}$ left of $L_2$.

$$u = -80 \, \text{cm}$$

Step 4: Apply the lens formula to $L_2$.

$$\frac{1}{v} - \frac{1}{u} = \frac{1}{f_2}$$ $$\frac{1}{v} - \frac{1}{-80} = \frac{1}{40}$$ $$\frac{1}{v} + \frac{1}{80} = \frac{1}{40}$$ $$\frac{1}{v} = \frac{1}{80}$$ $$v = 80 \, \text{cm}$$

Step 5: Convert this to distance from the concave lens.

$$x = 40 + 80 = 120 \, \text{cm}$$

Thus, the required numerical value is $120$.