Prism — Deviation, minimum deviation, dispersion
Ray Optics
12
JEE Qs
8%
Hard
75
min
Master the geometry and algebraic manipulation for prism deviation, as it's a frequently tested concept, especially the minimum deviation condition.
🧮 Key Formulas
✅ Key Points for JEE
- 1Always draw the ray path and correctly label angles (i, r1, r2, e, A) to apply Snell's law and geometric relations accurately.
- 2The angle of deviation (δ) depends on the angle of incidence (i), the angle of the prism (A), and the refractive index (n) of the prism material.
- 3The condition for minimum deviation (δ_min) significantly simplifies calculations: the incident and emergent rays are symmetrical with respect to the prism faces (i=e) and the refracted ray inside the prism is parallel to the base (r1=r2=A/2).
- 4Dispersion arises because the refractive index (n) of a material varies with the wavelength of light (n_violet > n_red); thus, violet light deviates more than red light.
- 5For thin prisms (prism angle A <= 10 degrees), the deviation and dispersion formulas simplify, allowing for quick approximations.
⚠️ Common Mistakes
- ✕Incorrectly applying Snell's Law at the second surface or misidentifying internal angles (r1, r2) with external ones (i, e).
- ✕Forgetting to consider the medium surrounding the prism; refractive indices are relative (n_prism / n_medium).
- ✕Confusing angular dispersion with deviation, or misusing the formulas for thin prisms when the prism is not thin.
- ✕Assuming minimum deviation condition applies to all problems; it's a specific scenario that simplifies the formulas.
📝 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, Chapter 9: Ray Optics and Optical Instruments