Solution & Explanation

### Related Formula

m approx fracLf_0 times fracDf_e

### Core Logic For a compound microscope in normal adjustment (image formed at infinity), the magnifying power is given by the standard approximation:

m simeq fracl Df_0 f_e

where

l

is the tube length,

D = 25text cm

is the least distance of distinct vision. ### Step 1: Substitute Values Given values:

l = 32text cm

f_0 = 2text cm

f_e = 4text cm

D = 25text cm

(standard assumption when not given)

m = frac322 times frac254

m = 16 times frac254 = 4 times 25 = 100

Microscope solution diagram for Q50 - JEE Main 2026 Morning

### Pattern Recognition Compound microscope formulas: Normal adjustment

to

image at

infty

m = (L/f_0)(D/f_e)

. Image at near point

D

m = (L/f_0)(1 + D/f_e)

. Default to standard approximation when given tube length. ### Evaluation Rubric / Model Answer null ### Chapter Mix Class 12 Physics: Ray Optics and Optical Instruments

More Ray Optics and Optical Instruments Previous-Year Questions — Page 7

Q15 jee_main_2025_24_jan_morning Lens Maker's Formula

A plano-convex lens having radius of curvature of first surface 2 cm exhibits focal length of

f_1

in air. Another plano-convex lens with first surface radius of curvature 3 cm has focal length of

f_2

when it is immersed in a liquid of refractive index 1.2. If both the lenses are made of same glass of refractive index 1.5, the ratio of

f_1

and

f_2

will be :-

A. 3:5
B. 1:3
C. 1:2
D. 2:3

Solution

### Related Formula Lens Maker's Formula for a lens in a surrounding medium of refractive index

mu_m

is:

frac1f = left(fracmu_textlensmu_m - 1 ight)left(frac1R_1 - frac1R_2 ight)

### Core Logic For a plano-convex configuration, the flat side has an infinite radius of curvature (

R_2 = infty implies frac1R_2 = 0

). ### Step 1: Evaluating Respective Focal Scales For the first lens setup in air (

mu_m = 1

) :

frac1f_1 = (1.5 - 1)left(frac12 - 0 ight) = 0.5 times frac12 = frac14 implies f_1 = 4text cm

For the second lens setup immersed inside fluid (

mu_m = 1.2

) :

frac1f_2 = left(frac1.51.2 - 1 ight)left(frac13 - 0 ight) = (1.25 - 1)frac13 = frac0.253 = frac112 implies f_2 = 12text cm

Taking their direct ratio :

f_1 : f_2 = 4 : 12 = 1 : 3

### Pattern Recognition Always separate the refractive index multiplier from the geometric shape factor. This lets you calculate each change independently before taking the final ratio. ### Evaluation Rubric / Model Answer null ### Chapter Mix Class 12 Physics: Ray Optics and Optical Instruments

Q6 jee_main_2025_28_jan_evening Refraction at Spherical Surfaces

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

from the vertex of the meniscus in A forms an image with a magnification of

-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

### Related Formula For refraction at a single spherical surface separating two mediums :

fracn_2v - fracn_1u = fracn_2 - n_1R

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 $v$ :

-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:

\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}$

Refraction at Spherical Surfaces diagram for Q6 - JEE Main 2025 Evening

R$ will mathematically return as a positive parameter. ### Evaluation Rubric / Model Answer null ### Chapter Mix Class 12 Physics: Ray Optics

Q13 jee_main_2025_28_jan_evening Reflection by Spherical Mirrors

A concave mirror produces an image of an object such that the distance between the object and image is

20 \, textcm

. If the magnification of the image is

-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

of a mirror is given by:

m = -fracvu

The mirror equation relates focal length to distance positions:

frac1f = frac1v + frac1u implies f = fracuvu+v

Radius of curvature

R = 2f

. ### Core Logic Given, magnification

m = -3

. This tells us the image is real and inverted[cite: 131, 757]:

-3 = -fracvu implies v = 3u

Since both real objects and real images lie on the same side in front of a concave mirror,

u

and

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

Therefore, object distance

u = -10 text cm

and image distance

v = -30 text cm

. Substitute into the focal equation formula:

f = frac(-10)(-30)-10 - 30 = frac300-40 = -7.5 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

tells you immediately that the object lies between the Focus (

F

) and Center of Curvature (

C

), while the image forms beyond

C

. This geometric layout instantly verifies that the radius value must exceed

10text cm

. ### Evaluation Rubric / Model Answer null ### Chapter Mix Class 12 Physics: Ray Optics

Q jee_main_2025_29_jan_morning Total Internal Reflection

At the interface between two materials having refractive indices

n_1

and

n_2

, the critical angle for reflection of an em wave is

theta_1C

. The

n_2

material is replaced by another material having refractive index

n_3

, such that the critical angle at the interface between

n_1

and

n_3

materials is

theta_2C

. If

n_3 > n_2 > n_1

;

fracn_2n_3 = frac25

and

sin theta_2C - sin theta_1C = frac12

, then

theta_1C

A. $sin^-1left(frac16n_1right)$
B. $sin^-1left(frac23n_1 ight)$ $\sin^{-1}\left(\frac{2}{3n_1} ight)$
C. $sin^-1left(frac56mathrmn_1 ight)$
D. $sin^-1left(frac13n_1 ight)$ $\sin^{-1}\left(\frac{1}{3n_1} ight)$

Solution

### Core Logic Critical angle setups dictate [cite: 613, 614]:

sin theta_1C = fracn_1n_2, quad sin theta_2C = fracn_1n_3

[cite: 613, 614] Given

sin theta_2C - sin theta_1C = frac12

[cite: 2, 615]:

fracn_1n_3 - fracn_1n_2 = frac12 implies n_1 left(frac25 - 1right) = fracn_22

[cite: 617, 619, 621]

fracn_1n_2 = -frac56

This produces an impossible negative value for a refractive index ratio, meaning the data contains a contradiction. This question was dropped by NTA.

Q jee_main_2025_29_jan_morning Lenses

Let

mathbfu

and

mathbfv

be the distances of the object and the image from a lens of focal length

f

. The correct graphical representation of

mathbfu

and

mathbfv

for a convex lens when

|mathbfu| > f

, is

Solution

### Related Formula

(u - f)(v - f) = f^2

> **Note:** Expanding

(u - f)(v - f) = f^2

yields

uv - f(u + v) = 0

, which simplifies directly to the magnitude form of the lens formula

frac1v + frac1u = frac1f

. --- ### Core Logic #### 1. Lens Formula and Sign Conventions For a convex lens forming a real image (

|u| > f

): * **Object distance (

u

):** Negative (

-u

) * **Image distance (

v

):** Positive (

+v

) * **Focal length (

f

):** Positive (

+f

) Substituting these into the standard lens formula

frac1v - frac1u = frac1f

frac1v - left(-frac1uright) = frac1f

frac1v + frac1u = frac1f

--- #### 2. Analyzing Graphical Behavior ##### **A. The

v

vs.

u

Graph** Rearranging the equation for

v

frac1v = frac1f - frac1u = fracu - fuf implies v = fracufu - f

* **When

u to f

:**

v to infty

(Vertical asymptote at

u = f

) * **When

u to infty

:**

v to f

(Horizontal asymptote at

v = f

) * **When

u = 2f

:**

v = 2f

This relationship represents a **rectangular hyperbola** when plotting magnitudes (or in the respective coordinate quadrants under Cartesian sign conventions). ##### **B. The

frac1v

vs.

frac1u

Graph** Plotting the reciprocals (

frac1v

on the

y

-axis and

frac1u

on the

x

-axis):

frac1v = -frac1u + frac1f

This matches the straight-line equation

y = mx + c

: * **Slope (

m

):**

-1

* **Intercept (

c

):**

frac1f

--- ### Chapter Mix Class 12 Physics: Ray Optics and Optical Instruments

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In a microscope the objective is having focal length $f_0 = 2text cm$ $f_{0} = 2\text{ cm}$ and eye-piece is having focal length $f_e = 4text cm$ $f_{e} = 4\text{ cm}$. The tube length is 32 cm. The magnification produced by this microscope for normal adjustment is ____.

Solution & Explanation

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In a microscope the objective is having focal length f_0 = 2text cm$f_{0} = 2\text{ cm}$ and eye-piece is having focal length f_e = 4text cm$f_{e} = 4\text{ cm}$. The tube length is 32 cm. The magnification produced by this microscope for normal adjustment is ____.

Solution & Explanation

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More Ray Optics and Optical Instruments Previous-Year Questions — Page 7

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In a microscope the objective is having focal length $f_0 = 2text cm$ $f_{0} = 2\text{ cm}$ and eye-piece is having focal length $f_e = 4text cm$ $f_{e} = 4\text{ cm}$. The tube length is 32 cm. The magnification produced by this microscope for normal adjustment is ____.