A bi-convex lens has radius of curvature of both the surfaces same as 1/6 cm. If this lens is required to be replaced…

Year	2026	2025	2024	Total
Questions	19	34	10	63

Year

2026

2025

2024

Total

Questions

Solution & Explanation

Related Formula

Lens Maker's Formula:

(1)/(f) = (μ - 1) ((1)/(R₁) - (1)/(R₂))

For a bi-convex lens of equal radii ( $R₁ = +R$ $R_1 = +R$, $R₂ = -R$ $R_2 = -R$):

(1)/(f) = (μ - 1) (2)/(R)

Power $P \propto (1)/(f)$ $P \propto \frac{1}{f}$.

Core Logic

For the initial lens:

$R = (1)/(6) cm$ $R = \frac{1}{6} \ \mathrm{cm}$
$(1)/(f₁) = (μ - 1) (2)/(1/6) = 12 (μ - 1)$

For the replacement lens ( $R₁ = +R₁$ $R_1 = +R_1$, $R₂ = -R₂$ $R_2 = -R_2$):

(1)/(f₂) = (μ - 1) ((1)/(R₁) + (1)/(R₂))

Since power must be preserved ( $f₁ = f₂$ $f_1 = f_2$):

(1)/(R₁) + (1)/(R₂) = (2)/(R) = 12 cm⁻¹

We need to find a combination where the sum of the reciprocals of the radii equals $12$ $12$.

Step 1: Verify the options

Let's check each choice:

Option (1): $R₁ = (1)/(3)$ $R_1 = \frac{1}{3}$, $R₂ = (1)/(3)$ $R_2 = \frac{1}{3}$:

(1)/(1/3) + (1)/(1/3) = 3 + 3 = 6 \neq 12

Option (2): $R₁ = (1)/(5)$ $R_1 = \frac{1}{5}$, $R₂ = (1)/(7)$ $R_2 = \frac{1}{7}$:

(1)/(1/5) + (1)/(1/7) = 5 + 7 = 12 (Correct Combination)

Option (3): $R₁ = (1)/(3)$ $R_1 = \frac{1}{3}$, $R₂ = (1)/(7)$ $R_2 = \frac{1}{7}$:

(1)/(1/3) + (1)/(1/7) = 3 + 7 = 10 \neq 12

Option (4): $R₁ = (1)/(6)$ $R_1 = \frac{1}{6}$, $R₂ = (1)/(9)$ $R_2 = \frac{1}{9}$:

(1)/(1/6) + (1)/(1/9) = 6 + 9 = 15 \neq 12

Thus, only Option (2) meets the physical conditions.

Pattern Recognition

Sees: Equivalent thin lens power with modified surfaces. Trap: Neglecting sign convention for the second spherical surface during substitution. Shortcut: If the radii are of the form $1/n$ $1/n$, then the sum of $n₁ + n₂$ $n_1 + n_2$ must equal $2 \times ninitial$ $2 \times n_{\text{initial}}$. Since initial $n = 6$ $n = 6$, $2 \times 6 = 12$ $2 \times 6 = 12$. The only pairing whose denominators add up to 12 is $5 + 7$ $5 + 7$.

Chapter Mix

Class 12 Physics: Ray Optics and Optical Instruments

More Ray Optics and Optical Instruments Previous-Year Questions — Page 3

Q47 jee_main_2026_23_january_evening Lenses

The size of the images of an object, formed by a thin lens are equal when the object is placed at two different positions 8 cm and 24 cm from the lens. The focal length of the lens is ____ cm.

Numerical Answer. Answer: 16 to 16

Solution

Related Formula

Magnification formula for thin lens in terms of $f$ $f$ and $u$ $u$:

m = (f)/(f + u)

Core Logic

Since the sizes of the images are equal at two different object positions but the focal length doesn't change, one image must be real and the other virtual. Hence, their magnifications are equal in magnitude but opposite in sign. $m₁ = -m₂$ $m_1 = -m_2$

Step 1: Set Up Magnification Equations

Object distance 1: $u₁ = -8 cm$ $u_1 = -8 \, \mathrm{cm}$ (virtual image case, closer to lens) Object distance 2: $u₂ = -24 cm$ $u_2 = -24 \, \mathrm{cm}$ (real image case)

Using proper sign convention:

(f)/(f + (-8)) = -( (f)/(f + (-24)) )

Step 2: Solve for Focal Length

(f)/(f - 8) = -(f)/(f - 24)

Assuming $f \neq 0$ $f \neq 0$:

f - 24 = -(f - 8)

$f - 24 = -f + 8$ $f - 24 = -f + 8$ $2f = 32$ $2f = 32$

f = 16 cm

Pattern Recognition

For a convex lens, if equal size images are formed at two different object distances $u₁$ $u_1$ and $u₂$ $u_2$, then $f = (u₁ + u₂)/(2)$ $f = \frac{u_1 + u_2}{2}$ if both distances are strictly positive magnitude values.

Chapter Mix

Class 12 Physics: Ray Optics and Optical Instruments

Q31 jee_main_2026_24_january_morning Prism

The exit surface of a prism with refractive index n is coated with a material having refractive index

(n)/(2)

. When this prism is set for minimum angle of deviation it exactly meets the condition of critical angle. The prism angle is ____

A. $60°$ $60^{\circ}$
B. $15°$ $15^{\circ}$
C. $30°$ $30^{\circ}$
D. $45°$ $45^{\circ}$

Solution

Related Formula

Minimum Deviation: r₁ = r₂ = (A)/(2)

Critical Angle: θc = (μ₂)/(μ₁)

Core Logic

For minimum deviation, $i = e$ $i = e$ and $r = (A)/(2)$ $r = \frac{A}{2}$. At the exit surface, the light ray exactly meets the condition of the critical angle.

For Total Internal Reflection (TIR) condition matching critical angle: $r = θc$ $r = \theta_c$

r = θc

r = (n/2)/(n) = (1)/(2)

Step 1: Solve for Prism Angle

Since $r = (1)/(2)$ $\sin r = \frac{1}{2}$, we have $r = 30°$ $r = 30^{\circ}$. Since $r = (A)/(2)$ $r = \frac{A}{2}$ at minimum deviation:

(A)/(2) = 30°

$A = 60°$ $A = 60^{\circ}$

Pattern Recognition

Critical angle at an interface is always $⁻¹(μᵣₐᵣₑ/μdense)$ . If $μᵣₐᵣₑ$ $\mu_{\text{rare}}$ is exactly half of $μdense$ $\mu_{\text{dense}}$, the critical angle is exactly $30^°$ $30^\circ$.

Chapter Mix

Class 12 Physics: Ray Optics and Optical Instruments

Q33 jee_main_2026_24_january_morning Optical Instruments

In a microscope of tube length 10 cm two convex lenses are arranged with focal length of 2 cm and 5 cm. Total magnification obtained with this system for normal adjustment is

(5)k

. The value of k is ____

A. $2$ $2$
B. $5$ $5$
C. $3.5$ $3.5$
D. $4$ $4$

Solution

Related Formula

M = ( )/(f₀) \cdot (D)/(fₑ)

Core Logic

Given parameters: Focal length of objective, $f₀ = 2 cm$ $f_0 = 2 \text{ cm}$ Focal length of eyepiece, $fₑ = 5 cm$ $f_e = 5 \text{ cm}$ Tube length, $= 10 cm$ $\ell = 10 \text{ cm}$ Least distance of distinct vision, $D = 25 cm$ $D = 25 \text{ cm}$

Step 1: Calculate Total Magnification

For normal adjustment (final image at infinity), the magnification formula is:

M = ( ( )/(f₀) ) ( (D)/(fₑ) )

M = ( (10)/(2) ) ( (25)/(5) ) = 5 \times 5 = 25

Since $M = 25 = 5²$ $M = 25 = 5^2$, and the problem states $M = (5)^k$ $M = (5)^k$, we compare to find $k=2$ $k=2$.

Pattern Recognition

For normal adjustment of a compound microscope, $M = mₒ \times mₑ \approx (L/fₒ) \times (D/fₑ)$ .

Chapter Mix

Class 12 Physics: Ray Optics and Optical Instruments

Q28 jee_main_2026_24_january_evening Lenses and Magnification

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

A. $Readings recorded by all persons are correct$
B. $Reading recorded by P₃ persons are incorrect$
C. $Reading recorded by P₃ and P₂ persons are incorrect$
D. $Reading recorded by P₄ and P₅ persons are incorrect$

Solution

Related Formula

P = 100f (in cm)

Core Logic

Given Power $P = +5D$ $P = +5\mathrm{D}$

(1)/(f) = 5 \Rightarrow f = 20 cm

If the object is between $f$ $f$ and $2f$ $2f$ (between $20cm$ $20\text{cm}$ and $40cm$ $40\text{cm}$), the image must be formed beyond $2f$ $2f$ (i.e., $v > 40cm$ $v > 40\text{cm}$) and it is magnified.

Step 1: Check P3 Reading

For $P₃$ $P_3$, $u = 35cm$ $u = 35\text{cm}$ (which is between $20cm$ $20\text{cm}$ and $40cm$ $40\text{cm}$). Thus, the image distance $v$ $v$ must be greater than $40cm$ $40\text{cm}$. However, the recorded reading is $v = 37cm$ $v = 37\text{cm}$, which is incorrect.

Pattern Recognition

For a convex lens, if $u$ $u$ lies between $f$ $f$ and $2f$ $2f$, $v$ $v$ must be greater than $2f$ $2f$. A simple boundary check on the $2f$ $2f$ mark ( $40cm$ $40\text{cm}$) reveals faulty readings instantly without solving the lens formula.

Chapter Mix

Class 12 Physics: Ray Optics and Optical Instruments

Q45 jee_main_2026_24_january_evening Mirrors and Magnification

Distance between an object and three times magnified real image is 40 cm. The focal length of the mirror used is ____ cm.

A. $- 15/2$ $- 15/2$
B. $- 10$ $- 10$
C. $- 20$ $- 20$
D. $- 15$ $- 15$

Solution

Related Formula

m = -(v)/(u)

(1)/(v) + (1)/(u) = (1)/(f)

Core Logic

For a real image, magnification $m$ $m$ is negative: $m = -3$ $m = -3$

-3 = -(v)/(u) v = 3u

Given the distance between object and image is $40 cm$ $40\text{ cm}$: For real images formed by a concave mirror, both $u$ $u$ and $v$ $v$ are on the same side. The distance between them is $|v| - |u| = 40 cm$ $|v| - |u| = 40\text{ cm}$.

Step 1: Determine Object and Image Distances

$3|u| - |u| = 40$ $3|u| - |u| = 40$

2|u| = 40 |u| = 20 cm

|v| = 60 cm

Applying sign convention:

u = -20 cm, v = -60 cm

Step 2: Compute Focal Length

(1)/(f) = (1)/(-60) + (1)/(-20)

(1)/(f) = (-1 - 3)/(60) = -(4)/(60)

f = -15 cm

Pattern Recognition

Real magnified image means concave mirror. The separation is $(m-1)|u|$ $(m-1)|u|$. Therefore $|u| = Δ / (m-1)$ $|u| = \Delta / (m-1)$. Then strictly apply $(1)/(f) = (1)/(v) + (1)/(u)$ with signs.

Chapter Mix

Class 12 Physics: Ray Optics and Optical Instruments

Solution & Explanation

Related Formula

Core Logic

Step 1: Verify the options

Pattern Recognition

Chapter Mix

Reference Study Guides

More Ray Optics and Optical Instruments Previous-Year Questions — Page 3

Solution

Related Formula

Core Logic

Step 1: Set Up Magnification Equations

Step 2: Solve for Focal Length

Pattern Recognition

Chapter Mix

Solution

Related Formula

Core Logic

Step 1: Solve for Prism Angle

Pattern Recognition

Chapter Mix

Solution

Related Formula

Core Logic

Step 1: Calculate Total Magnification

Pattern Recognition

Chapter Mix

Solution

Related Formula

Core Logic

Step 1: Check P3 Reading

Pattern Recognition

Chapter Mix

Solution

Related Formula

Core Logic

Step 1: Determine Object and Image Distances

Step 2: Compute Focal Length

Pattern Recognition

Chapter Mix

More Ray Optics and Optical Instruments Questions — jee_main_2025_02_april_evening

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