Optics Numericals for JEE Main: Complete Guide

Optics is one of the most reliable sources of marks in JEE Main physics, with 2–3 questions appearing across every session. While the chapter is conceptually manageable, the numericals require precise sign convention, careful formula identification, and fast arithmetic. Students who practice optics numericals systematically can solve each question in 90 seconds — a massive time advantage. This guide covers all major numerical types: mirror problems, lens combinations, refraction through slabs and prisms, and the wave optics calculations for interference and diffraction.

Mirror Numericals: Sign Convention Drills

All mirror and lens problems in JEE Main use the Cartesian sign convention: incident light travels left to right; distances measured from the pole/optical centre; right and upward are positive; left and downward are negative. For mirrors: 1/v + 1/u = 1/f, where f = R/2 (R = radius of curvature). Concave mirror: f is negative (focus in front = left side). Convex mirror: f is positive. Magnification m = −v/u. If m is negative, image is inverted (real); if positive, image is erect (virtual). Classic JEE numerical: object at 30 cm in front of a concave mirror of radius 40 cm — find image position and type. Solution: u = −30 cm, f = −20 cm. 1/v = 1/f − 1/u = 1/(−20) − 1/(−30) = −3/60 + 2/60 = −1/60. So v = −60 cm (real, in front of mirror). m = −(−60)/(−30) = −2 (inverted, magnified). For a broader optics study resource, see our Ray and Wave Optics Guide.

Virtual image in a concave mirror occurs when the object is between the focus and pole (u less than f). In this case, v comes out positive — virtual image behind the mirror. This region is tested in JEE Main with questions like: "A candle is placed between F and C of a concave mirror. Describe the image." Answer: real, inverted, magnified, beyond C. Learn all six standard object positions for concave mirrors and the resulting image properties — a 30-second table recall during the exam can directly answer a conceptual question.

Lens Combination and Equivalent Focal Length

For two thin lenses in contact: 1/f_eq = 1/f1 + 1/f2. For lenses separated by distance d: 1/f_eq = 1/f1 + 1/f2 − d/(f1·f2). This formula generates multi-step JEE Main problems. Example: a convex lens (f = 30 cm) and a concave lens (f = −20 cm) are placed 10 cm apart. Find equivalent focal length. Solution: 1/f_eq = 1/30 + 1/(−20) − 10/(30×(−20)) = 1/30 − 1/20 + 10/600 = 20/600 − 30/600 + 10/600 = 0/600. Hmm, f_eq is infinity — a system with no focusing power (afocal). The same calculation with different separation values gives finite focal lengths — practice all variations. Solve lens combination problems on our JEE Main mock tests with instant answer validation and solution walkthrough.

Power of a lens in water: P_water = (n_lens/n_water − 1)(1/R1 − 1/R2). If a glass lens (n = 1.5) is submerged in water (n = 1.33), its focal length increases by a factor of approximately 4 compared to air. This is a frequently tested result — know the formula and the qualitative conclusion: a converging lens becomes less converging in water; if submerged in a medium denser than glass, it becomes diverging.

Wave Optics Numericals: Interference and Diffraction

Young's double slit experiment (YDSE): fringe width beta = lambda·D/d, where lambda is wavelength, D is screen distance, d is slit separation. Position of bright fringes: y_n = n·beta. Position of dark fringes: y_n = (n − 1/2)·beta. Intensity: I = 4·I_0·cos²(phi/2), where phi = (2·pi/lambda)·delta (path difference). For maximum intensity: delta = n·lambda. For zero intensity: delta = (n−1/2)·lambda. Standard JEE numerical: D = 1 m, d = 1 mm, lambda = 600 nm. Find: fringe width, position of 3rd bright fringe, minimum distance between a bright and an adjacent dark fringe. Fringe width = 600×10⁻⁹ × 1 / (1×10⁻³) = 0.6 mm. Position of 3rd bright = 3 × 0.6 = 1.8 mm. Distance between adjacent bright and dark = beta/2 = 0.3 mm.

Single slit diffraction: minima at a·sin(theta) = n·lambda, where a is slit width. Central maximum width (between first minima on both sides) = 2·lambda·D/a — twice the fringe width of YDSE with same geometry. The central maximum is brighter and wider than secondary maxima. JEE Main question: if the slit width in a single-slit experiment doubles, the central maximum width halves. The intensity at the centre: I_0 = (a/lambda)² × I_slit. Diffraction grating: d·sin(theta) = n·lambda gives principal maxima, where d is grating spacing. Know the resolving power R = nN, where N is total number of slits and n is order.

Prism Numericals and Optical Instruments

Minimum deviation through a prism: n = sin((A+D_m)/2)/sin(A/2). JEE Main standard: A = 60°, n = sqrt(3). Find D_m. sin(D_m/2 + 30°) = sqrt(3)×sin(30°) = sqrt(3)/2. So D_m/2 + 30° = 60°, D_m = 60°. This is a standard result: equilateral prism of glass (n = sqrt(3)) has minimum deviation of 60°. Lateral displacement through a rectangular glass slab: d = t·sin(i−r)/cos(r) where t is thickness. At normal incidence (i = 0), d = 0. Apparent depth formula revisited: for oblique observation at angle alpha, the apparent depth formula requires more complex analysis — but JEE Main almost always tests normal incidence. Register on our platform to access 150+ optics numerical problems with detailed solutions. Our premium plan gives access to chapter-specific practice sessions with difficulty calibration. For refraction at curved surfaces, which is foundational to all lens calculations, see our Refraction and Lenses Deep Dive Guide.

Final exam tip: in optics numericals, always write the sign convention explicitly in your working (especially u-negative for real objects) and check your final answer qualitatively before moving on. A positive image distance for a convex mirror is impossible if the object is real — catching this in 5 seconds prevents a wrong answer costing 4 marks.