Reflection — Laws, mirrors, mirror formula
Ray Optics
12
JEE Qs
8%
Hard
80
min
Master the New Cartesian sign convention and consistently apply it to all parameters (u, v, f, h_o, h_i) to avoid common errors in mirror formula applications.
🧮 Key Formulas
✅ Key Points for JEE
- 1The New Cartesian sign convention is crucial for correct application of mirror formula; always assume light travels from left to right, measure distances from the pole.
- 2The focal length (f) is positive for concave mirrors and negative for convex mirrors, assuming the convention where real focus is positive.
- 3Magnification (m) indicates image nature: |m|>1 (magnified), |m|<1 (diminished); m>0 (erect, virtual), m<0 (inverted, real).
- 4Real images are formed when reflected rays actually converge, always appear on the same side as the object (for concave mirrors, beyond focus), and can be projected on a screen. Virtual images are formed when reflected rays appear to diverge, always appear behind the mirror, and cannot be projected.
- 5For plane mirrors, f = infinity, R = infinity, m = +1, image is virtual, erect, and laterally inverted, with image distance equal to object distance.
⚠️ Common Mistakes
- ✕Incorrectly applying the New Cartesian sign convention for object distance (u), image distance (v), focal length (f), or height of object/image (h_o, h_i).
- ✕Algebraic errors when solving the mirror formula, especially with fractions and negative signs.
- ✕Confusing the properties of real vs. virtual images or erect vs. inverted images with their corresponding sign conventions.
📝 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