Practice Section
90 problems organized by difficulty - test your mastery
How to use: Start with Easy, move to Moderate only after 90%+ accuracy. Use timer for exam simulation.
Easy Level - CBSE Boards
Target: 90%+ accuracy in 30-40 seconds per question
An object is placed 40 cm from a concave mirror of focal length 15 cm. Find image distance and magnification.
u = -40 cm, f = -15 cm
1/v = 1/f - 1/u = -1/15 - 1/(-40) = -1/15 + 1/40 = (-8+3)/120 = -5/120 = -1/24
v = -24 cm (real image)
m = -v/u = -(-24)/(-40) = -0.6 (inverted, diminished)
A convex lens has focal length 20 cm. Object is at 30 cm. Find image position.
u = -30 cm, f = +20 cm
1/v = 1/f + 1/u = 1/20 + 1/(-30) = (3-2)/60 = 1/60
v = 60 cm (real image on opposite side)
Power of a lens is -2.5 D. Find focal length and type of lens.
P = 1/f → f = 1/P = 1/(-2.5) = -0.4 m = -40 cm
Negative focal length → Concave lens (diverging)
Critical angle for glass is 42°. Find refractive index of glass.
sin θc = 1/n
n = 1/sin(42°) = 1/0.669 ≈ 1.49 ≈ 1.5
A convex mirror has focal length 15 cm. Find radius of curvature.
R = 2f = 2 × 15 = 30 cm
25 more problems available in full practice set...
Moderate Level - NEET/JEE Main
Target: 80%+ accuracy in 60-90 seconds per question
Two thin lenses of focal lengths +10 cm and -20 cm are in contact. Find power of combination.
P = P₁ + P₂ = 1/0.1 + 1/(-0.2) = 10 - 5 = 5 D
Or: 1/f = 1/10 + 1/(-20) = (2-1)/20 = 1/20 → f = 20 cm → P = 5 D
A prism has refracting angle 60°. Minimum deviation is 30°. Find refractive index.
n = sin[(A+δm)/2] / sin(A/2)
n = sin[(60+30)/2] / sin(60/2) = sin(45°)/sin(30°) = (√2/2)/(1/2) = √2 ≈ 1.414
A simple microscope has focal length 5 cm. Find magnifying power when image is at near point (D = 25 cm).
m = 1 + D/f = 1 + 25/5 = 1 + 5 = 6
Convex lens forms real image 3 times object size. If focal length is 15 cm, find object and image distances.
m = v/u = 3 (real image, inverted, so m = -3 actually)
v = -3u (taking sign into account properly: v/u = -3 for inverted)
1/15 = 1/v - 1/u = 1/(-3u) - 1/u = -4/(3u)
u = -20 cm, v = 60 cm
An astronomical telescope has objective focal length 100 cm and eyepiece 5 cm. Find magnifying power in normal adjustment.
m = -fo/fe = -100/5 = -20 (magnitude = 20)
35 more problems available in full practice set...
Advanced Level - JEE Advanced
Target: 70%+ accuracy in 3-5 minutes per question
A lens made of material with RI 1.5 has focal length 20 cm in air. It is immersed in liquid of RI 1.2. Find new focal length.
In air: 1/20 = (1.5/1 - 1)(1/R₁ - 1/R₂) = 0.5(1/R₁ - 1/R₂)
In liquid: 1/f' = (1.5/1.2 - 1)(1/R₁ - 1/R₂) = 0.25(1/R₁ - 1/R₂)
Divide: f'/20 = 0.25/0.5 = 0.5 → f' = 10 × 2 = 40 cm
Two lenses of focal length 20 cm each are separated by 60 cm. Object at 30 cm from first lens. Find position of final image.
First lens: 1/20 = 1/v₁ - 1/(-30) → v₁ = 60 cm
This image is at same position as second lens (60 cm from first)
For second lens: Object is AT the lens (u₂ = 0?) - Special case!
Actually, u₂ = -(60-60) = 0 means object at optical center → image at infinity
A concave mirror and convex lens are placed coaxially 20 cm apart. Parallel beam incident on lens emerges parallel after reflection from mirror. If f_lens = 15 cm, find f_mirror.
Lens focuses parallel rays at 15 cm (its focus)
This point is 20-15 = 5 cm from mirror
For rays to emerge parallel after reflection, this must be at mirror's focus
Therefore: f_mirror = -5 cm (negative for concave)
17 more advanced problems available...
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