Human Eye — Defects and correction
Ray Optics
12
JEE Qs
8%
Hard
60
min
Master the application of the thin lens formula with correct sign conventions for each defect type, especially for far point and near point distances.
🧮 Key Formulas
✅ Key Points for JEE
- 1**Myopia (Nearsightedness)**: The eye's far point is finite (closer than infinity). Distant objects appear blurry. Image forms in front of the retina. Corrected using a **concave (diverging) lens**. For calculation, the lens forms a virtual image of an object at infinity (u = -infinity) at the person's far point (v = -far_point).
- 2**Hypermetropia (Farsightedness)**: The eye's near point is beyond 25 cm. Near objects appear blurry. Image forms behind the retina. Corrected using a **convex (converging) lens**. For calculation, the lens forms a virtual image of an object at 25 cm (normal near point, u = -25 cm) at the person's defective near point (v = -near_point).
- 3**Presbyopia**: Age-related loss of accommodation (elasticity of ciliary muscles and lens), primarily affecting near vision. Often corrected with bifocal lenses (upper concave for distance, lower convex for near).
- 4**Astigmatism**: Defect where the cornea or lens has different curvatures in different planes, causing light from a point source to focus as a line. Corrected using **cylindrical lenses** oriented appropriately.
- 5Always apply proper Cartesian sign conventions for object distance (u), image distance (v), and focal length (f). The far point and near point for the defective eye should be taken as image distances (v) and are always negative (virtual image on the same side as the object).
⚠️ Common Mistakes
- ✕Confusing the corrective lenses for myopia (concave) and hypermetropia (convex) or mixing up their characteristics (far point vs. near point issues).
- ✕Incorrect application of sign conventions, especially for 'u' and 'v' in the lens formula, leading to errors in focal length or power calculations.
- ✕Not understanding that 'far point' is the maximum distance at which a person can see clearly, and 'near point' is the minimum, and applying them incorrectly as 'u' or 'v' in the lens formula.
📝 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