In a long glass tube, mixture of two liquids A and B with refractive indices 1.3 and 1.4 respectively, forms a convex refractive meniscus towards A. If an object placed at 13mathrmcm$13\mathrm{cm}$ from the vertex of the meniscus in A forms an image with a magnification of -2$-2$ then the radius of curvature of meniscus is :
A.1 \, textcm$1 \, \text{cm}$
B.frac13mathrmcm$\frac{1}{3}\mathrm{cm}$
C.frac23 mathrm~cm$\frac{2}{3} \mathrm{~cm}$
D.frac43 mathrm~cm$\frac{4}{3} \mathrm{~cm}$
Solution & Explanation
### Related Formula
For refraction at a single spherical surface separating two mediums :
fracn_2v - fracn_1u = fracn_2 - n_1R$\frac{n_2}{v} - \frac{n_1}{u} = \frac{n_2 - n_1}{R}$
Linear magnification for a spherical refracting boundary is given by:
m = fracv / n_2u / n_1 = fracv cdot n_1u cdot n_2$
### Core Logic
Given parameters from the text [cite: 17, 651, 654]:
* Refractive index of Medium A, $n_1 = 1.3$
* Refractive index of Medium B, $n_2 = 1.4$
* Object distance, $u = -13 text cm$
* Magnification, $m = -2$
Using the magnification formula to locate image position $v$ :
$m = \frac{v / n_2}{u / n_1} = \frac{v \cdot n_1}{u \cdot n_2}$
### Core Logic
Given parameters from the \text [cite: 17, 651, 654]:
* Refractive index of Medium A, $n_1 = 1.3$
* Refractive index of Medium B, $n_2 = 1.4$
* Object distance, $u = -13 \text{ cm}$
* Magnification, $m = -2$
Using the magnification formula to locate image position $v$ :
$-2 = \frac{v \cdot 1.3}{(-13) \cdot 1.4}$
$-2 = \frac{1.3 \cdot v}{-18.2} \implies 1.3 v = 36.4 \implies v = 28 \text{ cm}
Now substitute $u = -13$, $v = 28$, $n_1 = 1.3$, $n_2 = 1.4$ into the boundary equation:
$
Now substitute $u = -13$, $v = 28$, $n_1 = 1.3$, $n_2 = 1.4$ into the boundary equation:
$\frac{1.4}{28} - \frac{1.3}{-13} = \frac{1.4 - 1.3}{R}$
$\frac{1}{20} + \frac{1}{10} = \frac{0.1}{R}$
$\frac{1 + 2}{20} = \frac{0.1}{R} \implies \frac{3}{20} = \frac{1}{10R}$
$30 R = 20 \implies R = \frac{2}{3} \text{ cm}$
### Step 1: Visual Context
The visual system configuration of the refracting interface is tracked here:
Refraction at Spherical Surfaces diagram for Q6 - JEE Main 2025 Evening
### Pattern Recognition
Keep precise track of sign conventions for single spherical surfaces. A convex meniscus towards A means the center of curvature lies inside medium B, meaning $
### Step 1: Visual Context
The visual system configuration of the refracting interface is tracked here:
Refraction at Spherical Surfaces diagram for Q6 - JEE Main 2025 Evening
### Pattern Recognition
Keep precise track of sign conventions for single spherical surfaces. A convex meniscus towards A means the center of curvature lies inside medium B, meaning $R$ will mathematically return as a positive parameter.
### Evaluation Rubric / Model Answer
null
### Chapter Mix
Class 12 Physics: Ray Optics
Keywords:#refraction at spherical surface formula#JEE Main 2025 Evening Q6#Ray Optics JEE Main 2025#radius of curvature of meniscus
More Ray Optics Previous-Year Questions — Page 2
Q132025Reflection by Spherical Mirrors
A concave mirror produces an image of an object such that the distance between the object and image is 20 \, textcm$20 \, \text{cm}$. If the magnification of the image is -3'$-3'$ , then the magnitude of the radius of curvature of the mirror is:
A.3.75mathrmcm$3.75\mathrm{cm}$
B.30mathrmcm$30\mathrm{cm}$
C.7.5mathrmcm$7.5\mathrm{cm}$
D.15mathrmcm$15\mathrm{cm}$
Solution
### Related Formula
Magnification m$m$ of a mirror is given by:
m = -fracvu$m = -\frac{v}{u}$
The mirror equation relates focal length to distance positions:
frac1f = frac1v + frac1u implies f = fracuvu+v$\frac{1}{f} = \frac{1}{v} + \frac{1}{u} \implies f = \frac{uv}{u+v}$
Radius of curvature R = 2f$R = 2f$.
### Core Logic
Given, magnification m = -3$m = -3$. This tells us the image is real and inverted[cite: 131, 757]:
-3 = -fracvu implies v = 3u$-3 = -\frac{v}{u} \implies v = 3u$
Since both real objects and real images lie on the same side in front of a concave mirror, u$u$ and v$v$ are both negative fields. The physical distance separation between them is:
|v| - |u| = 20 implies 3|u| - |u| = 20 implies 2|u| = 20 implies |u| = 10 text cm$|v| - |u| = 20 \implies 3|u| - |u| = 20 \implies 2|u| = 20 \implies |u| = 10 \text{ cm}$
Therefore, object distance u = -10 text cm$u = -10 \text{ cm}$ and image distance v = -30 text cm$v = -30 \text{ cm}$ .
Substitute into the focal equation formula:
f = frac(-10)(-30)-10 - 30 = frac300-40 = -7.5 text cm$f = \frac{(-10)(-30)}{-10 - 30} = \frac{300}{-40} = -7.5 \text{ cm}$R = 2 times |f| = 2 times 7.5 = 15 text cm$R = 2 \times |f| = 2 \times 7.5 = 15 \text{ cm}$
### Step 1: Visual Diagram
The visual positioning profile tracking focal path boundaries is given below:
Reflection by Spherical Mirrors diagram for Q13 - JEE Main 2025 Evening
### Pattern Recognition
A magnification of -3$-3$ tells you immediately that the object lies between the Focus (F$F$) and Center of Curvature (C$C$), while the image forms beyond C$C$. This geometric layout instantly verifies that the radius value must exceed 10text cm$10\text{ cm}$.
### Evaluation Rubric / Model Answer
null
### Chapter Mix
Class 12 Physics: Ray Optics
More Ray Optics Questions — jee_main_2025_28_jan_evening
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