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)

Solution & Explanation

### 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

Reference Study Guides

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

Q34 jee_main_2024_31_jan_morning Prism Deviation
The refractive index of a prism with apex angle A is cot(A/2). The angle of minimum deviation is :
  • A. delta_mathrmm = 180^circ - A
  • B. delta_mathrmm = 180^circ - 3A
  • C. delta_mathrmm = 180^circ - 4A
  • D. delta_mathrmm = 180^circ - 2A

Solution

### Related Formula mu = fracsinleft(fracA + delta_m2right)sinleft(fracA2right) ### Core Logic Given that the refractive index mu = cotleft(fracA2right). Substituting this into the prism formula: cotleft(fracA2right) = fracsinleft(fracA + delta_m2right)sinleft(fracA2right) fraccosleft(fracA2right)sinleft(fracA2right) = fracsinleft(fracA + delta_m2right)sinleft(fracA2right) Equating the numerators: cosleft(fracA2right) = sinleft(fracA + delta_m2right) We can rewrite cosine in terms of sine: sinleft(fracpi2 - fracA2right) = sinleft(fracA + delta_m2right) ### Step 2: Solve for Deviation Comparing the angles inside the sine functions: fracpi2 - fracA2 = fracA2 + fracdelta_m2 Multiply the entire equation by 2: pi - A = A + delta_m delta_m = pi - 2A Converting radians to degrees: delta_m = 180^circ - 2A ### Pattern Recognition Whenever refractive index mu = cot(A/2), the relation sin(90^circ - A/2) strictly matches the prism sine equation, meaning minimum deviation delta_m is always 180^circ - 2A. ### Evaluation Rubric / Model Answer null ### Chapter Mix Class 12 Physics: Ray Optics And Optical Instruments

More Ray Optics and Optical Instruments Questions — jee_main_2025_07_april_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%)