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

Solution & Explanation

### 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
Ray Optics combination diagram
Ray Optics combination diagram
### 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

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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

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