Q.26 The minimum light intensity that the human eye can perceive is 10−10 W m−210−10 W m−2. The area of opening of our eye (the pupil) is approximately 0.4 cm20.4 cm2. Consider yellow light with wavelength $\lambda =600\text{nm}$. The number of photons incident on the retina per second at the minimum intensity the eye can respond is ____.

The correct number of photons incident on the retina per second is ≈1.2 × 10⁴, so option (D) is correct.

Introduction

The minimum light intensity human eye photons per second problem is a classic numerical in modern physics and competitive exams, testing understanding of intensity, photon energy, and basic unit conversions.

By analysing intensity on the pupil area and dividing by energy per photon, one can estimate how many photons must reach the retina every second for the eye to just respond.

Step‑by‑step solution

Given data

• Minimum light intensity, I = 10⁻¹⁰ W m⁻².
• Area of pupil, A = 0.4 cm² = 0.4 × 10⁻⁴ m² = 4.0 × 10⁻⁵ m².
• Wavelength of yellow light, λ = 600 nm = 600 × 10⁻⁹ m = 6.0 × 10⁻⁷ m.
• Planck’s constant, h ≈ 6.63 × 10⁻³⁴ J s.
• Speed of light, c ≈ 3.0 × 10⁸ m s⁻¹.

1. Power falling on the pupil

Intensity is power per unit area, I = P/A.

So power on the eye:

P = I × A = 10⁻¹⁰ × 4.0 × 10⁻⁵ = 4.0 × 10⁻¹⁵ W.

This means 4.0 × 10⁻¹⁵ J falls on the pupil each second at the threshold of vision.

2. Energy of one photon at 600 nm

Energy of a photon is

E_photon = hc/λ.

Substituting values:

E_photon = (6.63 × 10⁻³⁴ × 3.0 × 10⁸) / (6.0 × 10⁻⁷) ≈ 3.3 × 10⁻¹⁹ J.

This matches standard calculations for 600 nm photons.

3. Number of photons per second

The number of photons per second, n, is total energy per second divided by energy per photon:

n = P / E_photon = (4.0 × 10⁻¹⁵) / (3.3 × 10⁻¹⁹) ≈ 1.2 × 10⁴.

So about 12,000 photons per second must reach the retina at the minimum visible intensity.

Therefore, the correct option is (D) 1.2 × 10⁴.

Explanation of each option

The computed value n ≈ 1.2 × 10⁴ photons s⁻¹ is the benchmark to evaluate all options.

• (A) 1.5 × 10³ – This is an order of magnitude smaller than the calculated value (about 10 times less) and would require either a much smaller intensity or a much larger photon energy than given, so it is inconsistent with the data.
• (B) 5 × 10³ – This option is closer but still roughly a factor of 2–3 lower than the correct result, suggesting an error such as using the wrong area or a poorly rounded photon energy.
• (C) 8 × 10³ – This value may arise from approximations in intermediate steps, but it is still significantly below the more accurate estimate 1.2 × 10⁴.
• (D) 1.2 × 10⁴ – This matches the accurate calculation using proper unit conversion and the formula E = hc/λ, hence it is the correct choice.

Key takeaways for exam preparation

• Convert all quantities to SI units (m² for area, m for wavelength) before applying formulas in photon‑intensity problems.
• Use P = I × A to find total power on the detector, then divide by E_photon = hc/λ to get the number of photons per second, a pattern that recurs in many competitive exam questions.