Physics is where JEE ranks are usually won or lost, since a single conceptual slip changes the whole answer. Questions here span mechanics, rotational motion, thermodynamics, waves and oscillations, electrostatics, current electricity, magnetism, optics, and modern physics. Numerical problems carry full derivations so you can see which step you skipped, not just which option was right.
Resolving power = 1/(1.22λ/2NA). It depends on wavelength and numerical aperture (NA = n×sin(θ))
Q.5Medium
When white light passes through a prism, violet light deviates more than red light. This is because:
Answer: D
Higher frequency → higher refractive index → slower speed in medium → greater deviation. All statements are correct.
Q.6Medium
An object is placed 10 cm from a concave lens of focal length 20 cm. Find the magnification.
Answer: C
Using lens formula: 1/(-20) = 101 + 1/v. v = -6.67 cm. Magnification m = -v/u = 6.1067 = 0.67
Q.7Medium
Polarization of light proves that light is:
Answer: B
Only transverse waves can be polarized. Polarization demonstrates the transverse nature of electromagnetic waves.
Q.8Medium
The intensity at a point in the interference pattern of two coherent sources is I₁ and I₂. The resultant intensity is maximum when the phase difference is:
Answer: B
Maximum intensity occurs for constructive interference when phase difference = 0 or 2π. I_max = (√I₁ + √I₂)²
Q.9Medium
An object is placed at distance u from a convex lens of focal length f. If the magnification is -2, what is the relationship between u and f?
Answer: A
Magnification m = -v/u = -2, so v = 2u. Using lens equation: 1/f = 1/u + 1/v = 1/u + 1/(2u) = 3/(2u). Therefore u = 3f/2.
Q.10Medium
In Young's double-slit experiment with slit separation d = 1 mm and distance to screen D = 1 m, if the 5th bright fringe is at 2.5 mm from the center, what is the wavelength of light?
Answer: A
For bright fringes: y = (m·λ·D)/d. For 5th bright fringe: 2.5 × 10⁻³ = (5 × λ × 1)/(1 × 10⁻³). Therefore λ = 500 nm.
Q.11Medium
A ray undergoes total internal reflection at a critical angle θc. If the refractive index of the denser medium is √2, what is θc?
Answer: B
At critical angle: sin(θc) = 1/n = 1/√2. Therefore θc = 45°. This occurs when light travels from denser to less dense (rarer) medium.
Q.12Medium
A concave lens of focal length -20 cm is used to form an image of an object placed 10 cm from it. What is the nature of the image?
Answer: B
For concave lens, images are always virtual, erect, and diminished regardless of object position. Using 1/v = 1/f - 1/u = -201 - 101 = -203, v = -320 ≈ -6.67 cm (virtual).
Q.13Medium
In a Newton's rings experiment, the diameter of the 10th dark ring is 0.5 cm. What is the diameter of the 5th dark ring?
A ray of light is incident on a glass slab at 60°. If the refractive index of glass is √3, what is the angle of refraction?
Answer: A
Using Snell's law: sin(60°) = √3 × sin(r). √23 = √3 × sin(r). sin(r) = 21, therefore r = 30°.
Q.15Medium
A prism has apex angle A = 60° and refractive index n = √3. What is the minimum angle of deviation?
Answer: A
At minimum deviation: A = r₁ + r₂ = 2r (by symmetry). Also, sin(A/2) = n·sin(r/2). sin(30°) = √3·sin(30°), which checks out. δ_m = 2i - A where i = A/2 + δ_m/2. Solving: δ_m = 30°.
Q.16Medium
An object moves towards a concave mirror of focal length 15 cm. Initially at 30 cm, it moves to 20 cm. How does the magnification change?
Answer: A
At u = 30 cm: m = -f/(u-f) = -1515 = -1. At u = 20 cm: m = -515 = -3. Magnification increases in magnitude from 1 to 3.
Q.17Medium
In an optical fiber, light undergoes total internal reflection. If the core has n = 1.5 and cladding has n = 1.48, what is the critical angle inside the core?
A lens combination has two lenses with powers P₁ = +10 D and P₂ = +5 D placed in contact. What is the focal length of the combination?
Answer: A
For lenses in contact: P_total = P₁ + P₂ = 10 + 5 = 15 D. Therefore f = 1/P = 151 ≈ 0.067 m = 6.7 cm.
Q.19Medium
A convex lens of power 5 diopters is placed at 15 cm from a plane mirror. An object is kept at 30 cm from the lens (on the opposite side of mirror). What is the position of final image?
Answer: A
Focal length f = 1/P = 51 = 0.2 m = 20 cm. For object at 30 cm: 1/f = 1/v + 1/u gives 201 = 1/v + 301, so v = 60 cm. Mirror acts at 15 cm, creating a complex system requiring stepwise analysis leading to final image at 30 cm.
Q.20Medium
In a single slit diffraction pattern, the first minimum occurs at an angle of 30°. If the slit width is doubled, at what angle will the first minimum occur?
Answer: A
For single slit diffraction, first minimum: a·sin(θ) = λ. If slit width is doubled, 2a·sin(θ') = λ, so sin(θ') = sin(θ)/2. Since sin(30°) = 0.5, sin(θ') = 0.25, therefore θ' ≈ 15°.
