Five persons P1 , P2 , P3 , P4 and P5 recorded object distance (u) and image distance (v) using same convex lens having power +5D as (25,96), (30,62), (35,37), (45,35) and (50,32) respectively. Identify correct statement

Given Power P = +5D If the object is between f and 2f (between 20cm and 40cm), the image must be formed beyond 2f (i.e., v > 40cm) and it is magnified. For P3, u = 35cm (which is between 20cm and 40cm). Thus, the image distance v must be greater than 40cm. However, the recorded reading is v = 37cm, which is incorrect. For a convex lens, if u lies between f and 2f, v must be greater than 2f. A simple boundary check on the 2f mark (40cm) reveals faulty readings instantly without solving the lens formula. Class 12 Physics: Ray Optics and Optical Instruments

Five persons P1 , P2 , P3 , P4 and P5 recorded object distance (u) and image distance (v) using same convex lens having power +5D as (25,96), (30,62), (35,37), (45,35) and (50,32) respectively. Identify correct statement

Given Power P = +5D If the object is between f and 2f (between 20cm and 40cm), the image must be formed beyond 2f (i.e., v > 40cm) and it is magnified. For P3, u = 35cm (which is between 20cm and 40cm). Thus, the image distance v must be greater than 40cm. However, the recorded reading is v = 37cm, which is incorrect. For a convex lens, if u lies between f and 2f, v must be greater than 2f. A simple boundary check on the 2f mark (40cm) reveals faulty readings instantly without solving the lens formula. Class 12 Physics: Ray Optics and Optical Instruments

Five persons P1 , P2 , P3 , P4 and P5 recorded object distance (u) and image distance (v) using same convex lens having…

Year	2026	2025	2024	Total
Questions	19	34	10	63

Q jee_main_2025_24_jan_morning Silvering of Lenses

A thin plano convex lens made of glass of refractive index 1.5 is immersed in a liquid of refractive index 1.2. When the plane side of the lens is silver coated for complete reflection, the lens immersed in the liquid behaves like a concave mirror of focal length 0.2 m. The radius of curvature of the curved surface of the lens is :-

A. 0.15 m
B. 0.10 m
C. 0.20 m
D. 0.25 m

Solution

Related Formula

The net focal power of a silvered tracking lens system is given by:

P = 2PL + PM

(1)/(f) = 2fL + 1fM

Core Logic

As shown in diagram

Silvering of Lenses diagram for Q12 - JEE Main 2025 Morning

, the plane flat side boundary interface has an infinite radius of curvature (

R₂ = \infty

), meaning its mirror focal component is

fM = \infty PM = 0

. The power depends entirely on the refraction step:

(1)/(f) = 2fL

Step 1: Lens Maker Formulation

Find the focal expression of the immersed lens element [cite: 91, 679]:

1fL = ( μglassμliquid - 1)((1)/(R)) = ((1.5)/(1.2) - 1)(1)/(R) = (0.3)/(1.2)(1)/(R) = (1)/(4R)

Now insert this into the total system tracking balance relation :

(1)/(f) = 2 ((1)/(4R)) = (1)/(2R)

Given the final effective concave configuration matches $f = 0.2 m$ $f = 0.2\text{ m}$ :

(1)/(0.2) = (1)/(2R) 2R = 0.2 R = 0.10 m

Pattern Recognition

Silvering a plano-flat back boundary means light traverses the initial curved face interface exactly twice, mapping to $R = 2 \cdot f \cdot (μrel - 1)$ .

Chapter Mix

Class 12 Physics: Ray Optics and Optical Instruments

Q 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

f₁

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

f₂

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

f₁

and

f₂

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 $μm$ $\mu_{m}$ is:

(1)/(f) = ( μlensμm - 1)( 1R₁ - 1R₂)

Core Logic

For a plano-convex configuration, the flat side has an infinite radius of curvature ( $R₂ = \infty 1R₂ = 0$ ).

Step 1: Evaluating Respective Focal Scales

For the first lens setup in air ( $μm = 1$ $\mu_{m} = 1$) :

1f₁ = (1.5 - 1)((1)/(2) - 0) = 0.5 \times (1)/(2) = (1)/(4) f₁ = 4 cm

For the second lens setup immersed inside fluid ( $μm = 1.2$ $\mu_{m} = 1.2$) :

1f₂ = ((1.5)/(1.2) - 1)((1)/(3) - 0) = (1.25 - 1)(1)/(3) = (0.25)/(3) = (1)/(12) f₂ = 12 cm

Taking their direct ratio :

f₁ : f₂ = 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.

Chapter Mix

Class 12 Physics: Ray Optics and Optical Instruments

Q2 jee_main_2025_24_jan_morning Power of a Lens

What is the relative decrease in focal length of a lens for an increase in optical power by 0.1 D from 2.5 D ? ['D' stands for dioptre]

A. 0.04
B. 0.40
C. 0.1
D. 0.01

Solution

Related Formula

The relationship between optical power $P$ $P$ and focal length $F$ $F$ is given by:

F = (1)/(P)

Relative decrease in focal length is defined as:

(Δ F)/(F) = (F - F')/(F)

Core Logic

Given initial power $P = 2.5 D$ $P = 2.5\text{ D}$[cite: 19, 600]. After an increase of $0.1 D$ $0.1\text{ D}$, the new power is[cite: 19, 603]:

P' = 2.5 + 0.1 = 2.6 D

Step 1: Calculate Focal Length Change

Find the initial and final focal lengths [cite: 602, 604]:

F = (1)/(2.5) = (2)/(5)

F' = (1)/(2.6) = (5)/(13)

Now, calculate the relative decrease:

(F - F')/(F) = 1 - (F')/(F) = 1 - (P)/(P') = 1 - (2.5)/(2.6) = (0.1)/(2.6) = (1)/(26) \approx 0.04

Pattern Recognition

For a small change, we can approximate using differentiation: $P = (1)/(F) dP = -(dF)/(F²) (dF)/(F) = -(dP)/(P)$ . Thus, the relative change magnitude is $(0.1)/(2.5) = (1)/(25) = 0.04$ $\frac{0.1}{2.5} = \frac{1}{25} = 0.04$.

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

13cm

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. $1 cm$ $1 \, \text{cm}$
B. $(1)/(3)cm$ $\frac{1}{3}\mathrm{cm}$
C. $(2)/(3) ~cm$ $\frac{2}{3} \mathrm{~cm}$
D. $(4)/(3) ~cm$ $\frac{4}{3} \mathrm{~cm}$

Solution

Related Formula

For refraction at a single spherical surface separating two mediums :

(n₂)/(v) - (n₁)/(u) = (n₂ - n₁)/(R)

Linear magnification for a spherical refracting boundary is given by:

$m = (v / n₂)/(u / n₁) = (v \cdot n₁)/(u \cdot n₂)$$

Core Logic

Given parameters from the [cite: 17, 651, 654]:

Refractive index of Medium A, $n₁ = 1.3$
Refractive index of Medium B, $n₂ = 1.4$
Object distance, $u = -13 cm$
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 $u = -13$, $v = 28$, $n₁ = 1.3$, $n₂ = 1.4$ 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

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 cm

. If the magnification of the image is

-3'

, then the magnitude of the radius of curvature of the mirror is:

A. $3.75cm$ $3.75\mathrm{cm}$
B. $30cm$ $30\mathrm{cm}$
C. $7.5cm$ $7.5\mathrm{cm}$
D. $15cm$ $15\mathrm{cm}$

Solution

Related Formula

Magnification $m$ $m$ of a mirror is given by:

m = -(v)/(u)

The mirror equation relates focal length to distance positions:

(1)/(f) = (1)/(v) + (1)/(u) f = (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 = -(v)/(u) 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 3|u| - |u| = 20 2|u| = 20 |u| = 10 cm

Therefore, object distance $u = -10 cm$ $u = -10 \text{ cm}$ and image distance $v = -30 cm$ $v = -30 \text{ cm}$ .

Substitute into the focal equation formula:

f = ((-10)(-30))/(-10 - 30) = (300)/(-40) = -7.5 cm

R = 2 \times |f| = 2 \times 7.5 = 15 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 $10 cm$ $10\text{ cm}$.

Chapter Mix

Class 12 Physics: Ray Optics

Five persons P₁$P_{1}$ , P₂$P_{2}$ , P₃$P_{3}$ , P₄$P_{4}$ and P₅$P_{5}$ recorded object distance (u) and image distance (v) using same convex lens having power +5D as (25,96), (30,62), (35,37), (45,35) and (50,32) respectively. Identify correct statement

Solution & Explanation

Related Formula

Core Logic

Step 1: Check P3 Reading

Pattern Recognition

Chapter Mix

Reference Study Guides

More Ray Optics and Optical Instruments Previous-Year Questions — Page 10

Solution

Related Formula

Core Logic

Step 1: Lens Maker Formulation

Pattern Recognition

Chapter Mix

Solution

Related Formula

Core Logic

Step 1: Evaluating Respective Focal Scales

Pattern Recognition

Chapter Mix

Solution

Related Formula

Core Logic

Step 1: Calculate Focal Length Change

Pattern Recognition

Chapter Mix

Solution

Related Formula

Core Logic

Core Logic

Step 1: Visual Context

Pattern Recognition

Step 1: Visual Context

Pattern Recognition

Chapter Mix

Solution

Related Formula

Core Logic

Step 1: Visual Diagram

Pattern Recognition

Chapter Mix

More Ray Optics and Optical Instruments Questions — jee_main_2026_24_january_evening

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Five persons $P₁$ $P_{1}$ , $P₂$ $P_{2}$ , $P₃$ $P_{3}$ , $P₄$ $P_{4}$ and $P₅$ $P_{5}$ recorded object distance (u) and image distance (v) using same convex lens having power +5D as (25,96), (30,62), (35,37), (45,35) and (50,32) respectively. Identify correct statement