GEOMETRICAL OPTICS_JEE-Advanced 2025-S2 - Competishun – Live JEE & NEET Courses and Exam Prep

JEE ADVANCED 2025

Paper-2 2025

Multiple correct answers - Select all that apply

Question

Two identical concave mirrors each of focal length f are facing each other as shown in the schematic diagram. The focal length f is much larger than the size of the mirrors. A glass slab of thickness t and refractive index n0 is kept equidistant from the mirrors and perpendicular to their common principal axis. A monochromatic point light source S is embedded at the center of the slab on the principal axis, as shown in the schematic diagram. For the image to be formed on S itself, which of the following distances between the two mirrors is/are correct:

Select ALL correct options:

$4 f+\left(1-\frac{1}{n_0}\right) t$

$2 f+\left(1-\frac{1}{n_0}\right) t$

$4 f+\left(n_0-1\right) t$

$2 f+\left(n_0-1\right) t$

✓ Correct! Well done.

✗ Incorrect. Try again or view the solution.

⚠ Partially correct. Some answers are missing.

Solution

\begin{aligned} & h=\frac{h_0}{\mu_0} \\ & \Rightarrow d+\frac{t}{2 n_0}=2 f \\ & d=2 f-\frac{t}{2 n_0} \end{aligned}

$\begin{aligned} & d+\frac{t}{2}=2 f-\frac{t}{2 n_0}+\frac{t}{2} \\ & 2\left(d+\frac{t}{2}\right)=4 f+\frac{t}{2}\left(1-\frac{1}{n_0}\right) \\ & =4 f+t\left(1-\frac{1}{n_0}\right) \\ & d+\frac{t}{2 n_0}=f \\ & d=f-\frac{t}{2 n_0} \\ & d+\frac{t}{2}=f-\frac{t}{2 n_0}+\frac{t}{2} \\ & 2\left(d+\frac{t}{2}\right)=2 f+t\left(1-\frac{1}{n_0}\right)\end{aligned}$

Question Tags

JEE Advance

Physics

Easy

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