A biconvex lens of refractive index 1.5 has a focal length of 20 mathrm~cm$20 \mathrm{~cm}$ in air. Its focal length when immersed in a liquid of refractive index 1.6 will be:
A.-16 mathrm~cm$-16 \mathrm{~cm}$
B.-160 mathrm~cm$-160 \mathrm{~cm}$
C.+160 mathrm~cm$+160 \mathrm{~cm}$
D.+16 mathrm~cm$+16 \mathrm{~cm}$
Solution & Explanation
### Related Formula
From the Lens Maker's Formula:
frac1f = left( fracmu_textlensmu_textmedium - 1 right) left( frac1R_1 - frac1R_2 right)$\frac{1}{f} = \left( \frac{\mu_{\text{lens}}}{\mu_{\text{medium}}} - 1 \right) \left( \frac{1}{R_1} - \frac{1}{R_2} \right)$
Taking the ratio of focal length in liquid medium (f_m$f_m$) to focal length in air (f_a$f_a$):
fracf_mf_a = frac(mu_1 - 1) mu_mmu_1 - mu_m$\frac{f_m}{f_a} = \frac{(\mu_1 - 1) \mu_m}{\mu_1 - \mu_m}$
where,
mu_1$\mu_1$ = refractive index of the lens material = 1.5
mu_m$\mu_m$ = refractive index of the liquid medium = 1.6
f_a$f_a$ = focal length in air = 20 mathrm~cm$20 \mathrm{~cm}$
### Core Logic
Substitute the parameters into the relative ratio template equation:
fracf_m20 = frac(1.5 - 1) times 1.61.5 - 1.6$\frac{f_m}{20} = \frac{(1.5 - 1) \times 1.6}{1.5 - 1.6}$
### Step 1: Simplify and Compute
$fracf_m20 = frac0.5 times 1.6-0.1$$\frac{f_m}{20} = \frac{0.5 \times 1.6}{-0.1}$fracf_m20 = frac0.8-0.1 = -8$\frac{f_m}{20} = \frac{0.8}{-0.1} = -8$f_m = -8 times 20 = -160 mathrm~cm$f_m = -8 \times 20 = -160 \mathrm{~cm}$
Therefore, the focal length in the liquid is -160 mathrm~cm$-160 \mathrm{~cm}$.
### Pattern Recognition
Notice that since the surrounding liquid medium has a higher refractive index than the lens material itself (mu_m gt mu_1$\mu_m \gt \mu_1$), the sign of the focal length flips from positive to negative. The convex lens behaves as a diverging lens inside this specific liquid.
### Evaluation Rubric / Model Answer
null
### Chapter Mix
Class 12 Physics: Ray Optics and Optical Instruments
Keywords:#biconvex lens immersed in a liquid#JEE Main 2024 Morning Q45#Optics JEE Main 2024#Refraction at Spherical Surfaces and by Lenses
More Ray Optics and Optical Instruments Previous-Year Questions — Page 6
Q14jee_main_2025_24_jan_eveningLenses and Magnification
A photograph of a landscape is captured by a drone camera at a height of 18 km. The size of the camera film is 2 \, cm times 2 \, cm$2 \, cm \times 2 \, cm$ and the area of the landscape photographed is 400 \, km^2$400 \, km^2$ . The focal length of the lens in the drone camera is:
A. 1.8 cm
B. 2.8 cm
C. 2.5 cm
D. 0.9 cm
Solution
### Related Formula
Areal Magnification:
m^2 = fracA_textimageA_textobject = left(fracff+u
ight)^2 approx left(fracfu
ight)^2$m^2 = \frac{A_{\text{image}}}{A_{\text{object}}} = \left(\frac{f}{f+u}
ight)^2 \approx \left(\frac{f}{u}
ight)^2$
since object distance u = -18\ mathrmkm$u = -18\ \mathrm{km}$ is vastly larger than f$f$.
### Core Logic
Given parameters: Ray context geometry for drone camera scaling layout Q14
- Object height distance, H = 18\ mathrmkm = 18 times 10^3\ mathrmm$H = 18\ \mathrm{km} = 18 \times 10^3\ \mathrm{m}$
- Film size area, A_textimage = 2\ mathrmcm times 2\ mathrmcm = 4\ mathrmcm^2 = 4 times 10^-4\ mathrmm^2$A_{\text{image}} = 2\ \mathrm{cm} \times 2\ \mathrm{cm} = 4\ \mathrm{cm^2} = 4 \times 10^{-4}\ \mathrm{m^2}$
- Landscape area, A_textobject = 400\ mathrmkm^2 = 400 times 10^6\ mathrmm^2$A_{\text{object}} = 400\ \mathrm{km^2} = 400 \times 10^6\ \mathrm{m^2}$
Linear magnification factor:
fracyx = sqrtfracA_textimageA_textobject = sqrtfrac4 times 10^-4400 times 10^6 = sqrt10^-12 = 10^-6$\frac{y}{x} = \sqrt{\frac{A_{\text{image}}}{A_{\text{object}}}} = \sqrt{\frac{4 \times 10^{-4}}{400 \times 10^6}} = \sqrt{10^{-12}} = 10^{-6}$
Using the simple pinhole/thin lens perspective ratio:
fracfH = 10^-6 implies f = 18 times 10^3 times 10^-6 = 18 times 10^-3\ mathrmm = 1.8\ mathrmcm$\frac{f}{H} = 10^{-6} \implies f = 18 \times 10^3 \times 10^{-6} = 18 \times 10^{-3}\ \mathrm{m} = 1.8\ \mathrm{cm}$
### Pattern Recognition
For aerial satellite imaging contexts where u gg f$u \gg f$, linear sizing scales directly as fractextfilm sidetextground side = fracfH$\frac{\text{film side}}{\text{ground side}} = \frac{f}{H}$.
### Evaluation Rubric / Model Answer
null
### Chapter Mix
Class 12 Physics: Ray Optics and Optical Instruments
Q2jee_main_2025_24_jan_morningPower 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 = frac1P$F = \frac{1}{P}$Relative decrease in focal length is defined as:
fracDelta FF = fracF - F'F$\frac{\Delta F}{F} = \frac{F - F'}{F}$
### Core Logic
Given initial power P = 2.5text D$P = 2.5\text{ D}$[cite: 19, 600]. After an increase of 0.1text D$0.1\text{ D}$, the new power is[cite: 19, 603]:
P' = 2.5 + 0.1 = 2.6text D$P' = 2.5 + 0.1 = 2.6\text{ D}$
### Step 1: Calculate Focal Length Change
Find the initial and final focal lengths [cite: 602, 604]:
F = frac12.5 = frac25$F = \frac{1}{2.5} = \frac{2}{5}$F' = frac12.6 = frac513$F' = \frac{1}{2.6} = \frac{5}{13}$
Now, calculate the relative decrease:
fracF - F'F = 1 - fracF'F = 1 - fracPP' = 1 - frac2.52.6 = frac0.12.6 = frac126 approx 0.04$\frac{F - F'}{F} = 1 - \frac{F'}{F} = 1 - \frac{P}{P'} = 1 - \frac{2.5}{2.6} = \frac{0.1}{2.6} = \frac{1}{26} \approx 0.04$
### Pattern Recognition
For a small change, we can approximate using differentiation: P = frac1F implies dP = -fracdFF^2 implies fracdFF = -fracdPP$P = \frac{1}{F} \implies dP = -\frac{dF}{F^2} \implies \frac{dF}{F} = -\frac{dP}{P}$. Thus, the relative change magnitude is frac0.12.5 = frac125 = 0.04$\frac{0.1}{2.5} = \frac{1}{25} = 0.04$.
### Evaluation Rubric / Model Answer
null
### Chapter Mix
Class 12 Physics: Ray Optics and Optical Instruments
Q12jee_main_2025_24_jan_morningSilvering 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 = 2P_L + P_M$P = 2P_{L} + P_{M}$frac1f = frac2f_L + frac1f_M$\frac{1}{f} = \frac{2}{f_{L}} + \frac{1}{f_{M}}$
### 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 (R_2 = infty$R_{2} = \infty$), meaning its mirror focal component is f_M = infty implies P_M = 0$f_{M} = \infty \implies P_{M} = 0$. The power depends entirely on the refraction step:
frac1f = frac2f_L $\frac{1}{f} = \frac{2}{f_{L}} $
### Step 1: Lens Maker Formulation
Find the focal expression of the immersed lens element [cite: 91, 679]:
frac1f_L = left(fracmu_textglassmu_textliquid - 1
ight)left(frac1R
ight) = left(frac1.51.2 - 1
ight)frac1R = frac0.31.2frac1R = frac14R$\frac{1}{f_{L}} = \left(\frac{\mu_{\text{glass}}}{\mu_{\text{liquid}}} - 1
ight)\left(\frac{1}{R}
ight) = \left(\frac{1.5}{1.2} - 1
ight)\frac{1}{R} = \frac{0.3}{1.2}\frac{1}{R} = \frac{1}{4R}$
Now insert this into the total system tracking balance relation :
frac1f = 2 left(frac14R
ight) = frac12R $\frac{1}{f} = 2 \left(\frac{1}{4R}
ight) = \frac{1}{2R} $
Given the final effective concave configuration matches f = 0.2text m$f = 0.2\text{ m}$ :
frac10.2 = frac12R implies 2R = 0.2 implies R = 0.10text m$\frac{1}{0.2} = \frac{1}{2R} \implies 2R = 0.2 \implies R = 0.10\text{ m}$
### 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 (mu_textrel - 1)$R = 2 \cdot f \cdot (\mu_{\text{rel}} - 1)$.
### Evaluation Rubric / Model Answer
null
### Chapter Mix
Class 12 Physics: Ray Optics and Optical Instruments
Q15jee_main_2025_24_jan_morningLens Maker's Formula
A plano-convex lens having radius of curvature of first surface 2 cm exhibits focal length of f_1$f_{1}$ in air. Another plano-convex lens with first surface radius of curvature 3 cm has focal length of f_2$f_{2}$ 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$f_{1}$ and f_2$f_{2}$ 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$\mu_{m}$ is:
frac1f = left(fracmu_textlensmu_m - 1
ight)left(frac1R_1 - frac1R_2
ight)$\frac{1}{f} = \left(\frac{\mu_{\text{lens}}}{\mu_{m}} - 1
ight)\left(\frac{1}{R_{1}} - \frac{1}{R_{2}}
ight)$
### Core Logic
For a plano-convex configuration, the flat side has an infinite radius of curvature (R_2 = infty implies frac1R_2 = 0$R_{2} = \infty \implies \frac{1}{R_{2}} = 0$).
### Step 1: Evaluating Respective Focal Scales
For the first lens setup in air (mu_m = 1$\mu_{m} = 1$) :
frac1f_1 = (1.5 - 1)left(frac12 - 0
ight) = 0.5 times frac12 = frac14 implies f_1 = 4text cm$\frac{1}{f_{1}} = (1.5 - 1)\left(\frac{1}{2} - 0
ight) = 0.5 \times \frac{1}{2} = \frac{1}{4} \implies f_{1} = 4\text{ cm}$
For the second lens setup immersed inside fluid (mu_m = 1.2$\mu_{m} = 1.2$) :
frac1f_2 = left(frac1.51.2 - 1
ight)left(frac13 - 0
ight) = (1.25 - 1)frac13 = frac0.253 = frac112 implies f_2 = 12text cm$\frac{1}{f_{2}} = \left(\frac{1.5}{1.2} - 1
ight)\left(\frac{1}{3} - 0
ight) = (1.25 - 1)\frac{1}{3} = \frac{0.25}{3} = \frac{1}{12} \implies f_{2} = 12\text{ cm}$
Taking their direct ratio :
f_1 : f_2 = 4 : 12 = 1 : 3$f_{1} : f_{2} = 4 : 12 = 1 : 3$
### 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
Q6jee_main_2025_28_jan_eveningRefraction 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$13\mathrm{cm}$ 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
### 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
More Ray Optics and Optical Instruments Questions — jee_main_2024_29_jan_morning
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