Refraction at Spherical Surfaces
Ray Optics
12
JEE Qs
8%
Hard
75
min
Master the New Cartesian Sign Convention and practice its consistent application; it is the most critical aspect for solving problems accurately.
🧮 Key Formulas
✅ Key Points for JEE
- 1Strictly follow the New Cartesian Sign Convention: All distances are measured from the pole; distances measured in the direction of incident light are positive, opposite are negative; heights above the principal axis are positive, below are negative.
- 2n1 is the refractive index of the medium where the object is located (from which light originates), and n2 is the refractive index of the medium where the image is formed (into which light enters).
- 3The formula (n2/v) - (n1/u) = (n2 - n1)/R is universally applicable for all types of spherical refracting surfaces (convex or concave) and for both real and virtual objects/images.
- 4For a plane refracting surface, the radius of curvature R is taken as infinity, simplifying the formula to (n2/v) - (n1/u) = 0, or v = (n2/n1)*u for object in medium n1.
⚠️ Common Mistakes
- ✕Incorrectly applying the New Cartesian Sign Convention for object distance (u), image distance (v), or radius of curvature (R).
- ✕Swapping the refractive indices n1 and n2 in the main formula, leading to sign errors and incorrect results.
- ✕Not consistently using the same units for all distances (e.g., mixing cm and m) during calculations.
- ✕Misinterpreting the nature of the image (real/virtual) or its orientation (erect/inverted) based on the signs of v and magnification (m).
📝 Practice Questions
See allQ46.The driver sitting inside a parked car is watching vehicles approaching from behind with the help of his side view mirror, which is a convex mirror with radius of curvature R = 2 m . Another car approaches him from behind with a uniform speed of 90 km/hr. When the car is at a distance of 24 m from him, the magnitude of the acceleration of the image of the car in the side view mirror is ' a '. The value of 100 a is _______ m/s2 .
Q27.A spherical surface of radius of curvature R, separates air from glass (refractive index = 1.5 ). The centre of curvature is in the glass medium. A point object ' O ' placed in air on the optic axis of the surface, so that its real image is formed at ' I ' inside glass. The line OI intersects the spherical surface at P and PO = PI. The distance PO equals to (1) 5 R (2) 3 R (3) 1.5 R (4) 2 R
Q32.Given is a thin convex lens of glass (refractive index μ ) and each side having radius of curvature R. One side is polished for complete reflection. At what distance from the lens, an object be placed on the optic axis so that 2025 (22 Jan Shift 1) JEE Main Previous Year Paper the image gets formed on the object itself ? (1) R/μ (2) R/(2μ −3) (3) μR (4) R/(2μ −1)
Q42.In the diagram given below, there are three lenses formed. Considering negligible thickness of each of them as compared to |R1| and |R2|, i.e., the radii of curvature for upper and lower surfaces of the glass lens, the power of the combination is (1) 1 − + |R2| |R2| 6 ( |R1|1 1 ) (2) −16 ( |R1|1 1 ) (3) 1 + − |R2| |R2| 6 ( |R1|1 1 ) (4) −16 ( |R1|1 1 )
Q29.A symmetric thin biconvex lens is cut into four equal parts by two planes AB and CD as shown in figure. If the power of original lens is 4 D then the power of a part of the divided lens is (1) D (2) 8D (3) 2D (4) 4D
Q29.Given a thin convex lens (refractive index μ2 ), kept in a liquid (refractive index μ1, μ1 < μ2 ) having radii of curvatures |R1| and |R2|. Its second surface is silver polished. Where should an object be placed on the optic axis so that a real and inverted image is formed at the same place? (1) μ1|R1|⋅|R2| (2) μ1|R1|⋅|R2| μ2(|R1|+|R2|)−μ1|R2| μ2(|R1|+|R2|)−μ1|R1| (3) (μ2+μ1)|R1| (4) μ1|R1|⋅|R2| (μ2−μ1) μ2(2|R1|+|R2|)−μ1√|R1|⋅|R2|
NCERT Chapters
- Class 12 Physics Ch 9: Ray Optics and Optical Instruments