The minimum light intensity that can be perceived by eye is about 10-10 W m -2. The number of photons of wavelength 5.6 × 10-7 m that must enter the pupil of area 10-6 m 2 per second for vision is approximately (Take .h=6.6 × 10-34 J s -1)

Question

The minimum light intensity that can be perceived by eye is about 10-10 W m -2. The number of photons of wavelength 5.6 × 10-7 m that must enter the pupil of area 10-6 m 2 per second for vision is approximately (Take .h=6.6 × 10-34 J s -1) (A) 3 × 102 (B) 3 × 103 (C) 3 × 104 (D) 3 × 105

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Here, I=10-10 W m -2=10-10 J s -1 m -2 Let the number of photons that enter the pupil per second be N. Then (N h v/10-6)=10-10 or N=(10-6 × 10-10/h v)=(10-16 λ/h c) (∵ c=v λ) =(10-16 ×(5.6 × 10-7)/(6.6 × 10-34)(3 × 108)) =0.28 × 103 ≈ 3 × 102 photons / s

Q. The minimum light intensity that can be perceived by eye is about $1 0^{- 10} W m^{- 2}$ . The number of photons of wavelength $5.6 \times 1 0^{- 7} m$ that must enter the pupil of area $1 0^{- 6} m^{2}$ per second for vision is approximately (Take $h = 6.6 \times 1 0^{- 34} J s^{- 1})$

$3 \times 1 0^{2}$

$3 \times 1 0^{3}$

$3 \times 1 0^{4}$

$3 \times 1 0^{5}$

Q. The minimum light intensity that can be perceived by eye is about 10−10Wm−2. The number of photons of wavelength 5.6×10−7m that must enter the pupil of area 10−6m2 per second for vision is approximately (Take h=6.6×10−34Js−1)

3×102

3×103

3×104

3×105

Here, I=10−10Wm−2=10−10Js−1m−2 Let the number of photons that enter the pupil per second be N. Then 10−6Nhv​=10−10 or N=hv10−6×10−10​=hc10−16λ​ (∵c=vλ) =(6.6×10−34)(3×108)10−16×(5.6×10−7)​ =0.28×103≈3×102 photons / s

Q. The minimum light intensity that can be perceived by eye is about $1 0^{- 10} W m^{- 2}$ . The number of photons of wavelength $5.6 \times 1 0^{- 7} m$ that must enter the pupil of area $1 0^{- 6} m^{2}$ per second for vision is approximately (Take $h = 6.6 \times 1 0^{- 34} J s^{- 1})$

$3 \times 1 0^{2}$

$3 \times 1 0^{3}$

$3 \times 1 0^{4}$

$3 \times 1 0^{5}$