JEE Advanced 2023 — Paper 1 — PHY-III solution

JEE Advanced2023Paper 1PHY-III

An optical arrangement consists of two concave mirrors $M_{1}$ and $M_{2}$ , and a convex lens $L$ with a common principal axis, as shown in the figure. The focal length of $L$ is $10$ cm. The radii of curvature of $M_{1}$ and $M_{2}$ are $20$ cm and $24$ cm, respectively. The distance between $L$ and $M_{2}$ is $20$ cm. A point object $S$ is placed at the mid-point between $L$ and $M_{2}$ on the axis. When the distance between $L$ and $M_{1}$ is $n /7$ cm, one of the images coincides with $S$ . The value of $n$ is _______. (Assume that the light first refracts through L)

Your answer

Solution

Setup. Focal lengths: $f_{L} = 10$ cm (lens), $f_{M_{1}} = 10$ cm, $f_{M_{2}} = 12$ cm. The object $S$ sits $10$ cm to the left of $M_{2}$ and $10$ cm to the right of $L$ .

Path: light first refracts through $L$ , reflects off $M_{1}$ , refracts back through $L$ , reflects off $M_{2}$ , refracts through $L$ again, and arrives back at $S$ .

Step 1. Light from $S$ to the left strikes $L$ at distance $10$ cm. With $u = - 10$ , $f_{L} = 10$ , the lens equation gives $v = \infty$ ; the emerging beam is parallel.

Step 2. A parallel beam reflecting off $M_{1}$ converges to its focus, i.e. at distance $f_{M_{1}} = 10$ cm in front of $M_{1}$ .

Step 3. Let $d$ be the lens-to- $M_{1}$ distance. After reflection the image acts as object for $L$ at distance $(d - 10)$ to the left of $L$ . Refraction through $L$ : $\frac{1}{v} - \frac{1}{- ( d - 10 )} = \frac{1}{10}$ , giving $\frac{1}{v} = \frac{1}{10} - \frac{1}{d - 10}$ .

Step 4. The resulting image must serve as an object for $M_{2}$ whose final image, after another refraction through $L$ , coincides with $S$ . By tracing the return leg with $M_{2}$ producing an image at $10$ cm from $L$ , the consistency condition reduces to $v = 80$ cm (image to the right of $L$ acting as virtual object for $M_{2}$ ).

Step 5. Substituting in step 3: $\frac{1}{80} = \frac{1}{10} - \frac{1}{d - 10} \Rightarrow \frac{1}{d - 10} = \frac{1}{10} - \frac{1}{80} = \frac{7}{80}$ . Hence $d - 10 = \frac{80}{7}$ , giving $d = \frac{150}{7}$ cm.

Comparing with $d = n /7$ : $n = 150$ .

← Back to full paper

Practice more PHY-III

Concept-wise practice with instant solutions on Fundamenthol.

Start practicing →