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GATE 2026 IN – Question 65

Communication and Optical Instrumentation · Optical sources and detectors · 2 marks · Numerical answer

A photodiode has a responsivity of $0.8\ \mathrm{A/W}$. If the input optical power to the photodiode is 2 mW, the power delivered to a $50\ \Omega$ load is ________ μW. (Rounded off to the nearest integer.)

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Correct answer: 128 to 128

Explanation

The photocurrent is $I=\mathcal R P_{opt}=0.8\times0.002=0.0016\ \mathrm A$. Using the ideal photocurrent-source model, the load power is $P_L=I^2R=(0.0016)^2\times50=0.000128\ \mathrm W$.

Converting to microwatts gives **128 μW**.