Lens Formula + Prism Deviation
Ray Optics
55
JEE Qs
18%
Hard
75
min
Master sign conventions for lenses and practice geometric ray tracing for prisms to correctly apply formulas and analyze complex scenarios.
🧮 Key Formulas
✅ Key Points for JEE
- 1Consistent application of Cartesian sign convention is paramount for all lens formula calculations (object distance, image distance, focal length, radii of curvature).
- 2A real image is formed by actual intersection of refracted rays (v > 0 for lens), while a virtual image is formed by apparent intersection (v < 0 for lens).
- 3Minimum deviation in a prism occurs when the ray passes symmetrically through the prism, meaning the angle of incidence equals the angle of emergence (i = e) and the angles of refraction inside are equal (r₁ = r₂).
- 4The power of a lens (P=1/f) indicates its converging or diverging ability; convex lenses have positive power, concave lenses have negative power.
- 5For a given prism, the angle of deviation first decreases to a minimum and then increases as the angle of incidence increases.
⚠️ Common Mistakes
- ✕Incorrectly applying sign conventions for object/image distances, focal lengths, and radii of curvature in lens and lensmaker's formulas.
- ✕Confusing the lens formula (1/f = 1/v - 1/u) with the mirror formula (1/f = 1/v + 1/u), leading to sign errors.
- ✕Assuming minimum deviation conditions (i=e, etc.) are always applicable for any prism problem, instead of identifying specific conditions for general deviation.
- ✕Errors in basic trigonometric calculations or algebraic manipulation, especially when dealing with reciprocals or complex expressions for refractive index in prism problems.
📝 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 2 Ch 9: Ray Optics and Optical Instruments