A collimated beam of light of diameter 2 mm is propagating along x-axis. The beam is required to be expanded in a collimated beam of diameter 14 mm using a system of two convex lenses. If first lens has focal length 40 mm, then the focal length of second lens is ____ mm.

Numerical Answer Type:
Enter a numerical value Answer: 280 to 280 +4 marks

Solution & Explanation

### Related Formula textMagnification m = fracD_textoutD_textin = fracf_2f_1 ### Core Logic For a beam expander using two convex lenses, the lenses are arranged such that their focal points coincide. This creates an afocal system where parallel input rays remain parallel upon output.
Lenses solution diagram for Q46 - JEE Main 2026 Morning
Lenses solution diagram for Q46 - JEE Main 2026 Morning
By similar triangles at the focal point: tan theta = fracD_1 / 2f_1 = fracD_2 / 2f_2 ### Step 1: Solving for f2 fracD_1f_1 = fracD_2f_2 frac240 = frac14f_2 f_2 = 14 times frac402 = 14 times 20 = 280text mm ### Pattern Recognition A standard beam expander (Keplerian telescope design used in reverse) has f_2/f_1 = D_2/D_1. ### 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 3

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
  • B. 288mathrmmm
  • C. 555mathrmmm
  • D. 655mathrmmm

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
  • B. 1 mathrm~D
  • C. 20mathrmD
  • D. 10mathrmD

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
Q6 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
  • B. frac80011mathrm~cm
  • C. frac70011mathrm~cm
  • D. frac60011mathrm~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
  • B. 45
  • C. frac607
  • D. 15

Solution

### Related Formula $frac1f_texteq = frac1f_1 + frac1f_2 frac1v - frac1u = frac1f where, f_1 = focal length of the convex lens f_2 = focal length of the concave lens u = object distance v = image distance ### Core Logic Given parameters: - Convex lens: f_1 = +30\mathrm{~cm} - Concave lens: f_2 = -20\mathrm{~cm} - Object distance: u = -20\mathrm{~cm} (placed to the left of the lens system) First, find the equivalent focal length of the lens combination in contact: frac1f_texteq = frac130 + frac1-20 = frac2 - 360 = -frac160 implies f_texteq = -60mathrm~cm ### Step 1: Image Distance Calculation Use the thin lens formula: frac1v - frac1u = frac1f_texteq frac1v - frac1-20 = frac1-60 implies frac1v + frac120 = -frac160 frac1v = -frac160 - frac120 = frac-1 - 360 = -frac460 = -frac115 v = -15mathrm~cm ### Pattern Recognition Sees: Lenses in contact + object distance → Combination focal length first, then thin lens formula. Trap: Keep proper sign conventions. An object to the left implies u = -20\mathrm{~cm}. A negative image distance v = -15\mathrm{~cm} means a virtual image formed on the same side as the object. The question asks for "distance", which is the magnitude: |-15| = 15\mathrm{~cm}$. ✓ ### 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
  • B. 1:1
  • C. 2:1
  • D. 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

More Ray Optics and Optical Instruments Questions — jee_main_2026_21_jan_morning

Practice all Ray Optics and Optical Instruments previous-year questions →

Rankbit System
JEE Physics: Waves (+15.5%) | Electrostatics: Concentric Shells (-29.7%) | Modern Physics: Photoelectric Clones (+34.2%) | Mathematics: Definite Integrals (+18.1%) | Chemistry: Coordination Splitting (-11.4%) | JEE Physics: Waves (+15.5%) | Electrostatics: Concentric Shells (-29.7%) | Modern Physics: Photoelectric Clones (+34.2%) | Mathematics: Definite Integrals (+18.1%) | Chemistry: Coordination Splitting (-11.4%)
YOUR FIRST PREP STEP STARTS HERE

We Map Every Repeating Question in Competitive Exams.

Say goodbye to generic mock test fatigue. RankBit uses smart analysis to group past exam questions into their foundational Repeating Question Types. Find chapter weightage, track repeating questions, and score higher with targeted practice.

Select Your Target Exam

Choose an exam track below to find formulas per chapter and patterns.

Syncing Exam Intelligence

Mapping formulas and patterns across all tracks…

PATH A — FULL LENGTH PRACTICE

Full Mock Test Hub

Simulate real NTA exam conditions with fully tracked mocks. Time yourself against past papers.

Now Live Open
PATH B — TARGETED PRACTICE

Topic-wise Practice Hub

Practice past-year questions one chapter at a time. Pick an exam → subject → chapter and get every PYQ for that topic — pulled together from all past papers — with the chapter's key formulas alongside.

Loading Questions... Browse Topics

Exam Analysis Tools

Chapter weightage trends and topic-wise practice — all mapped from past shift papers.

Year-wise PYQ

Chapter Weightage & Trends

See high-yield chapters, year-wise counts and the frequency-based chance each topic repeats next exam.

Weightage Board Open
Targeted PYQs

Topic-wise Practice Hub

Every past-year question for a chapter pulled from all papers, with its key formulas alongside.

One chapter at a time Open
Latest from the Blog
View all →

Loading articles...