Convex lens and concave lens are in contact as shown in diagram
Power of convex lens P₁ = 1/+f₁ (in m) = 100/f₁ (in cm) = 100/+25 = 4D Power of convex lens P₂ = 1/+f₂ = 100/-20 = -5D Combined power = P₁ + P₂ = 4D + (-5D) = -1D System of lenses is diverging in nature.
(a) The IF stage is represented by box 'x' (intermediate frequency stage) Amplifier is represented by box 'y.' (b) The IF stage's function is to convert a high-frequency electromagnetic wave to a lower frequency for detection in the detector. Amplifier function - If the detected signal isn't powerful enough to be used, an amplifier is necessary. Its purpose is to boost the signal's power to the appropriate level.
When light with energy hn higher than energy gap Eg illuminates the junction of a diode (any type of diode), an electric field 'E' exists across the junction from n-side to p-side. Then, in or near the depletion area of the diode, photon absorption generates electron-hole pairs. Electrons and holes are separated due to the electric field. An emf is created when free electrons are gathered on the n-side and holes are captured on the p-side. An electric current of the order of μA flows through the external resistance as a result of the generated emf.
Detection of Optical Signals: Applying a reverse bias makes it easier to see how the current changes as the light intensity changes. As a result, a photodiode can be used to detect optical signals as a photodetector.
(i) The p-n junction is substantially doped. (ii) LEDs have relatively low reverse breakdown voltages. (iii) The semiconductor utilised to make visible LEDs must have a band gap of at least 1.8 eV. The order of band gap of an LED to emit light in the visible range is about 3 eV to 1.8 eV.
Q₂ =CV₂ .....(ii) From (i) and (ii) Q₁/Q₂ = CV₁/CV₂ => 360/120 = v/v-120 v = 180 volts Therefore, C = Q₁/V₁ = 360 x 10⁻⁶/180 = 2 x 10⁻⁶ C = 2 𝜇F (ii) If the voltage applied has increased by 120 V, then V₃ = 180 + 120 = 300 V Hence, the charge stored in the capacitor Q₃ = CV₃ = 2 x 10⁻⁶ x 300 = 600 𝜇C
(i) Electric flux through a surface φ = E(vector) . S(vector) Flux through the left surface φᴸ = -|E| |S| = -50 x. |S| Since x = 1m φᴸ = -50 x 1 x 25 x 10⁻⁴ = -1250 x 10⁻⁴ = - 0.125 Nm² C⁻¹ Flux through the right surface φᴿ = |E| |S| Since x = 2m, φᴿ = 50x |S| = 50 x 2 x 25 x 10⁻⁴ = 0.250 Nm² C⁻¹ Net flux through the cylinder φ net = φᴿ + φᴸ = 0.250 - 0.125 = 0.125 Nm² C⁻¹ (ii) Charge inside the cylinder, by Guass's theorem φ net = q/𝜀∘ = q = 𝜀∘φ net = 8.854 x 10⁻¹² x 0.125 = 8.854 x 10⁻¹² x 1/8 1.107 x 10⁻¹² C
