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#optical physics

5 public questions tagged with this topic.

What is the intensity of light in the photon picture, according to the wave optics concept?

**Intensity at point** in double-slit I = I_max cos²(φ/2), φ = (2π/λ)Δ, for Δ=λ/4 φ=π/2 I= I_max cos²(π/4)= I_max/2 =2I₀, for Δ=λ/3 φ=2π/3 I= I_max cos²(π/3)= I_max×0.25= I₀, for Δ=5λ/8 φ=5π/4? Actually φ=2π×5/8=5π/4, cos²(5π/8)=?, path difference for destructive φ=(2n+1)π, constructive 2nπ. In the photon picture, intensity is determined by the number of photons crossing a unit area per unit time. Using Δ = d sinθ, y = n λ D/d, a sinθ = n λ, I = I₀ cos²θ, n = c/v, sinC = 1/n, λ' = λ/n and A = 2a cos(φ/2), calculation gives Number of photons per unit area per unit time, illustrating interferenc

Ref: NCERT > Physics Book > Wave Optics > Superposition, Resultant Amplitude and Intensity

What is the maximum intensity of light transmitted through three polaroids when the first and third are crossed, and the

**Polaroid rotation** intensity varies sinusoidally with angle due to Malus law I = I₀ cos²θ, when polaroid rotated 90° from initial, intensity goes from max to zero, for unpolarized light rotating polaroid does not change intensity after first polaroid because average, but second polaroid intensity depends on relative angle, explains why intensity transmitted through two polaroids drops to zero when perpendicular. For crossed polaroids, intensity after the second polaroid is I₀ cos² θ , and after the third (at 90° - θ ) is I = I₀ cos² θ sin² θ = (I₀/4) sin² 2θ . Maximum occurs at θ = 45° , so

Ref: NCERT > Physics Book > Wave Optics > Polarization and Malus Law

In a convex lens, if an object is placed at the focal point, where is the image formed?

**Refraction at spherical surface** formula n₁/u + n₂/v = (n₂-n₁)/R governs single surface, extension to two surfaces yields lens maker. Double convex with equal |R| has f = R/[2(n-1)], for R=12 cm, n=1.5, f=12 cm, illustrating dependence on curvature and index. For a convex lens, when the object is at the focal point (F), the rays after refraction become parallel and do not converge to a point on the other side. The image is formed at infinity, as the rays appear to diverge from an infinitely distant point when traced backward. Substituting values gives At infinity, which matches expected ima

Ref: NCERT > Physics Book > Ray Optics > Refraction at Spherical Surfaces and Lens Maker's Formula

In a concave lens, what property of the lens determines the position of the virtual focal point?

**TIR application** requires n₁>n₂ and θ₁>C. Critical angle formula derived from Snell's law with θ₂=90°, n₁ sinC = n₂ sin90° = n₂, so sinC = n₂/n₁. For glass-air sinC=1/1.52, C≈41°, for water-air 48.6°, determining cutoff for transmission. The virtual focal point of a concave lens is where diverging rays appear to originate when traced backward. This position is determined by the lens’s focal length, which depends on its curvature and refractive index, defining the extent of divergence. Substituting values gives Focal length, which matches expected image position and magnification from mirror

Ref: NCERT > Physics Book > Ray Optics > Total Internal Reflection and Critical Angle

A prism of angle \( 40^\circ \) and refractive index \( 1.5 \) produces what minimum deviation?

**Prism minimum deviation** δ_m satisfies n = sin[(A+δ_m)/2]/sin(A/2), A prism angle (degrees), n refractive index, δ_m minimum deviation. For A=60°, n=1.45, sin[(60+δ_m)/2]=1.45×sin30°=0.725, (60+δ_m)/2=46.5°, 60+δ_m=93°, δ_m=33°, illustrating n increase raises δ_m. For a thin prism: D_m = (n - 1) A . n = 1.5 , A = 40° . D_m = (1.5 - 1) × 40 = 0.5 × 40 = 20° . Substituting values gives 20°, which matches expected image position and magnification from mirror/lens formula 1/f = 1/v - 1/u (lens) or 1/f = 1/v + 1/u (mirror), confirming sign conventions and refraction principles.

Ref: NCERT > Physics Book > Ray Optics > Refraction Through Prism and Minimum Deviation