Govt. Exams
Since I ∝ A², the ratio A₁/A₂ = √(I₁/I₂)
For closed pipe, f₁ = v/(4L) = 340/(4×0.5) = 340/2 = 170 Hz
Intensity ∝ A²f². New intensity = (2A)² × (f/2)² × I/(A²f²) = 4 × (1/4) × I = 2I
From y = Asin(2πx/λ - 2πt/T), A = 5 m, and f = 1/T = 1/2π × 2π = 1 Hz
For destructive interference, the path difference must be odd multiples of λ/2: (2n+1)λ/2 or (n+1/2)λ
Doppler effect: When source approaches observer, f' = f(v)/(v-vs), which is greater than f
Frequency remains constant during refraction. λ = v/f. Since speed increases from 330 to 1500 m/s, wavelength increases proportionally
In photoconductive (reverse-bias) mode, the depletion region widens significantly, reducing junction capacitance and transit time, resulting in faster response (MHz to GHz range) and shot noise is suppressed due to the reverse field.
The depletion width W = √(2εε₀(V_bi + V_R)/(eN_D N_A/(N_D + N_A))) depends on applied voltage, doping concentrations, and built-in potential. Both doping levels and reverse bias voltage are critical factors.
Maximum power point (MPP) occurs where P = V × I is maximum, typically at ~80% of V_oc and ~80% of I_sc, determined by the fill factor.
