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:

Solution & Explanation

### 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

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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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