Two identical objects are placed in front of convex mirror and concave mirror having same radii of curvature of 12 , cm…

Q8 jee_main_2025_07_april_morning Refraction at Spherical Surfaces and by Lenses

A lens having refractive index 1.6 has focal length of

12mathrmcm

, when it is in air. Find the focal length of the lens when it is placed in water. (Take refractive index of water as 1.28)

A. $355mathrmmm$ $355\mathrm{mm}$
B. $288mathrmmm$ $288\mathrm{mm}$
C. $555mathrmmm$ $555\mathrm{mm}$
D. $655mathrmmm$ $655\mathrm{mm}$

Solution

### Related Formula Lens Maker's Formula in a surrounding medium with refractive index

mu_m

is:

frac1f = left( fracmu_Lmu_m - 1 right) left( frac1R_1 - frac1R_2 right)

### Core Logic In air (

mu_m = 1

):

frac112 = (1.6 - 1) left( frac1R_1 - frac1R_2 right)

frac112 = 0.6 left( frac1R_1 - frac1R_2 right) implies left( frac1R_1 - frac1R_2 right) = frac112 times 0.6 = frac1072

### Step 1: Calculate Focal Length in Water In water (

mu_m = 1.28

):

frac1f_w = left( frac1.61.28 - 1 right) left( frac1072 right)

Simplify the relative index factor:

frac1.61.28 = frac160128 = 1.25

frac1f_w = (1.25 - 1) left( frac1072 right) = 0.25 times frac1072 = frac14 times frac1072 = frac10288

f_w = 28.8 mathrm~cm = 288 mathrm~mm

### Pattern Recognition Sees: Lens index

\mu_L = 1.6

, focal length in air

f_a

, and focal length in medium

f_m

. Shortcut: Use the ratio of focal lengths directly:

fracf_mf_a = fracmu_L - 1fracmu_Lmu_m - 1 = frac0.6frac1.61.28 - 1 = frac0.60.25 = 2.4

f_m = 2.4 times 12 = 28.8 mathrm~cm = 288 mathrm~mm

$ ### Evaluation Rubric / Model Answer null ### Chapter Mix Class 12 Physics: Ray Optics and Optical Instruments

Q10 jee_main_2025_07_april_morning Refraction at Spherical Surfaces and by Lenses

Two thin convex lenses of focal lengths

30~mathrmcm

and

10~mathrmcm

are placed coaxially,

10~mathrmcm

apart. The power of this combination is :

A. $5 mathrmD$ $5 \mathrm{D}$
B. $1 mathrm~D$ $1 \mathrm{~D}$
C. $20mathrmD$ $20\mathrm{D}$
D. $10mathrmD$ $10\mathrm{D}$

Solution

### Related Formula The equivalent focal length

f_texteq

of two thin coaxially aligned lenses separated by distance

d

is given by:

frac1f_texteq = frac1f_1 + frac1f_2 - fracdf_1 f_2

The equivalent power in diopters (

D

) when focal lengths are in meters is:

P = frac1f_texteq

### Core Logic Given parameters: -

f_1 = 30 mathrm~cm = 0.3 mathrm~m

-

f_2 = 10 mathrm~cm = 0.1 mathrm~m

-

d = 10 mathrm~cm = 0.1 mathrm~m

### Step 1: Calculate Power Substitute parameters into the equivalent focal length equation:

frac1f_texteq = frac10.3 + frac10.1 - frac0.10.3 times 0.1

frac1f_texteq = frac10.3 + 10 - frac10.3

frac1f_texteq = 10 mathrm~m^-1

P = 10 mathrm~D

### Pattern Recognition Sees: Lenses separated by distance

d

where

d = f_2

. Shortcut: Notice that

d = f_2 = 10 mathrm~cm

. When the separation distance between two thin lenses equals the focal length of the second lens, the equivalent power simplifies directly to

P = P_2 = 1/f_2 = 10 mathrm~D

since the terms

1/f_1

and

d/(f_1 f_2)

cancel out. ### Evaluation Rubric / Model Answer null ### Chapter Mix Class 12 Physics: Ray Optics and Optical Instruments

Q jee_main_2025_08_april_evening Refraction through Lenses

A concave-convex lens of refractive index

1.5

and the radii of curvature of its surfaces are

30mathrm~cm

and

20mathrm~cm

, respectively. The concave surface is upwards and is filled with a liquid of refractive index

1.3

. The focal length of the liquid-glass combination will be

A. $frac50011mathrm~cm$ $\frac{500}{11}\mathrm{~cm}$
B. $frac80011mathrm~cm$ $\frac{800}{11}\mathrm{~cm}$
C. $frac70011mathrm~cm$ $\frac{700}{11}\mathrm{~cm}$
D. $frac60011mathrm~cm$ $\frac{600}{11}\mathrm{~cm}$

Solution

### Related Formula

frac1f_texteq = frac1f_textliquid + frac1f_textglass

frac1f = (mu - 1)left(frac1R_1 - frac1R_2right)

### Core Logic Let's find the focal length of each individual lens in the combination: 1. **Liquid Lens**: - Refractive index,

mu_l = 1.3

- The upper surface is flat (exposed to air):

R_1 = infty

- The lower surface matches the upward concave surface of the glass lens:

R_2 = -30mathrm~cm

frac1f_textliquid = (1.3 - 1) left(frac1infty - frac1-30right) = 0.3 times frac130 = frac1100mathrm~cm^-1

2. **Glass Lens**: - Refractive index,

mu_g = 1.5

- First surface radius (concave upward),

R_1 = -30mathrm~cm

- Second surface radius (convex downward),

R_2 = -20mathrm~cm

(following light path downward)

frac1f_textglass = (1.5 - 1) left(frac1-30 - frac1-20right) = 0.5 left(-frac130 + frac120right) = 0.5 left(frac160right) = frac1120mathrm~cm^-1

### Step 1: Combination Focal Length Add the powers of both lenses:

frac1f_texteq = frac1100 + frac1120 = frac6 + 5600 = frac11600

f_texteq = frac60011mathrm~cm

### Pattern Recognition Sees: Glass lens with liquid poured on top → Think of it as a double lens system (liquid lens + glass lens). Trap: Be extremely careful with sign conventions for radii of curvature of the boundaries! Assume light travels from air through the liquid and then through the glass. This defines a consistent spatial propagation direction. ✓ ### Evaluation Rubric / Model Answer null ### Chapter Mix Class 12 Physics: Ray Optics

Q12 jee_main_2025_08_april_evening Refraction through Lenses

A convex lens of focal length

30mathrm~cm

is placed in contact with a concave lens of focal length

20mathrm~cm

. An object is placed at

20mathrm~cm

to the left of this lens system. The distance of the image from the lens in

mathrm~cm

is:

A. $30$ $30$
B. $45$ $45$
C. $frac607$ $\frac{60}{7}$
D. $15$ $15$

Solution

### Related Formula

$frac1f_texteq = frac1f_1 + frac1f_2

frac1v - frac1u = frac1f

where,

f_1

= focal length of the convex lens

f_2

= focal length of the concave lens

u

= object distance

v

= image distance ### Core Logic Given parameters: - Convex lens:

f_1 = +30\mathrm{~cm}

- Concave lens:

f_2 = -20\mathrm{~cm}

- Object distance:

u = -20\mathrm{~cm}

(placed to the left of the lens system) First, find the equivalent focal length of the lens combination in contact:

frac1f_texteq = frac130 + frac1-20 = frac2 - 360 = -frac160 implies f_texteq = -60mathrm~cm

### Step 1: Image Distance Calculation Use the thin lens formula:

frac1v - frac1u = frac1f_texteq

frac1v - frac1-20 = frac1-60 implies frac1v + frac120 = -frac160

frac1v = -frac160 - frac120 = frac-1 - 360 = -frac460 = -frac115

v = -15mathrm~cm

### Pattern Recognition Sees: Lenses in contact + object distance \to Combination focal length first, then thin lens formula. Trap: Keep proper sign conventions. An object to the left implies

u = -20\mathrm{~cm}

. A negative image distance

v = -15\mathrm{~cm}

means a virtual image formed on the same side as the object. The question asks for "distance", which is the magnitude:

|-15| = 15\mathrm{~cm}$. ✓ ### Evaluation Rubric / Model Answer null ### Chapter Mix Class 12 Physics: Ray Optics

Q7 jee_main_2025_29_jan_evening Cutting of Lenses

Two identical symmetric double convex lenses of focal length

f

are cut into two equal parts

L_1, L_2

by AB plane and

L_3, L_4

by XY plane as shown in figure respectively. The ratio of focal lengths of lenses

L_1

and

L_3

is:

Cutting of Lenses diagram for Q7 - JEE Main 2025 Evening — The figure details a convex lens being cut along the horizontal plane AB and vertical plane XY to create components L1, L2, L3, and L4.

A. $1:4$ $1:4$
B. $1:1$ $1:1$
C. $2:1$ $2:1$
D. $1:2$ $1:2$

Solution

### Related Formula

frac1f = (mu - 1)left(frac1R_1 - frac1R_2right)

### Core Logic 1. **Cutting along horizontal plane AB**: When a lens is cut along its principal axis, the radius of curvature of neither surface changes. Thus, the focal length of the split parts

L_1

and

L_2

remains exactly equal to the initial focal length:

f_L_1 = f

2. **Cutting along vertical plane XY**: When a lens is cut perpendicular to the principal axis into two symmetric plano-convex lenses, one surface becomes flat (

R_2 = infty

). According to Lens Maker's Formula, the focal length of parts

L_3

and

L_4

doubles:

f_L_3 = 2f

3. **Ratio Determination**:

fracf_L_1f_L_3 = fracf2f = frac12

Hence, the ratio is

1:2

. ### Pattern Recognition Shortcut rule for lens cutting: - Horizontal cut (along axis)

rightarrow

Focal length stays

f

. - Vertical cut (perp to axis)

rightarrow

Focal length doubles to

2f

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

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