Refraction — Snell's law, refractive index
Ray Optics
12
JEE Qs
8%
Hard
60
min
Always draw a clear ray diagram, correctly identify the normal at the point of incidence, and measure all angles with respect to the normal before applying Snell's Law.
🧮 Key Formulas
✅ Key Points for JEE
- 1Refraction is the bending of light as it passes from one medium to another due to a change in its speed.
- 2The frequency of light remains constant during refraction; only its speed and wavelength change.
- 3Light bends towards the normal when entering a denser medium (higher refractive index) and away from the normal when entering a rarer medium (lower refractive index).
- 4Snell's Law (n1 sinθ1 = n2 sinθ2) is the fundamental principle governing refraction, where angles (θ) are always measured with respect to the normal to the interface.
- 5For an object viewed normally from a rarer medium through a denser medium, its apparent depth is less than its real depth, given by n = real_depth / apparent_depth.
⚠️ Common Mistakes
- ✕Confusing the angle with the surface of incidence/refraction with the angle measured from the normal.
- ✕Incorrectly swapping n1 and n2 in Snell's law or the apparent depth formula, leading to inverted results.
- ✕Assuming the frequency of light changes during refraction, which is incorrect; only speed and wavelength change.
- ✕Failing to draw a clear ray diagram and correctly identify the normal at the point of incidence, especially for complex geometries.
📝 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 Part II Ch 9: Ray Optics and Optical Instruments