A parallel beam of light is incident on a fixed surface PQ at an angle 'θ' with the vertical as shown in the figure. The intensity of light is I and area of surface PQ is A. In List-I nature of reflection are given and in List-II force and generated radiation pressure are given. Then match List-I with suitable option List-II. (Given that I = 500 W/m2 , A=1.5m2,θ=370 )LIST-ILIST-IIP)If all the incident energy is absorbed by surface PQ then force on the surface and radiation pressure generated are respectively.1)3.2×10−6N,2.13×10−6N/m2Q)If all the incident energy is reflected by surface PQ then force on the surface and radiation pressure generated are respectively.2)2.24×10−6N,1.38×10−6N/m2R)If 30% incident energy is reflected by the surface then force on the surface and pressure generated are respectively. (Rest absorbed)3)2.74×10−6N,1.81×10−6N/m2S)If 70% incident energy is reflected back then force on the surface and radiation pressure are respectively. (Rest absorbed)4)2×10−6N,1.06×10−6N/m2 5)4.2×10−6N,3.13×10−6N/m2

Intensity is given by I=N(hCλ)(Acosθ)Δt,whereN→ total number of photons and λ→wavelength of photons (A) Fx=ΔPxΔt=IAcosθ.sinθC Fy=ΔPyΔt=IAcosθcosθC pr=FyA=Icos2θC (B) Fx=0 Fy=2IAcos2θC pr=2IAcos2θC(C) and (D) Fx=IAcos2θC(1−r) Fy=IAcos2θC(1+r) prIcos2θC(1+r)

A parallel beam of light is incident on a fixed surface PQ at an angle 'θ' with the vertical as shown in the figure. The intensity of light is I and area of surface PQ is A. In List-I nature of reflection are given and in List-II force and generated radiation pressure are given. Then match List-I with suitable option List-II. (Given that I = 500 W/m2 , A=1.5m2,θ=370 )LIST-ILIST-IIP)If all the incident energy is absorbed by surface PQ then force on the surface and radiation pressure generated are respectively.1)3.2×10−6N,2.13×10−6N/m2Q)If all the incident energy is reflected by surface PQ then force on the surface and radiation pressure generated are respectively.2)2.24×10−6N,1.38×10−6N/m2R)If 30% incident energy is reflected by the surface then force on the surface and pressure generated are respectively. (Rest absorbed)3)2.74×10−6N,1.81×10−6N/m2S)If 70% incident energy is reflected back then force on the surface and radiation pressure are respectively. (Rest absorbed)4)2×10−6N,1.06×10−6N/m2 5)4.2×10−6N,3.13×10−6N/m2

Intensity is given by I=N(hCλ)(Acosθ)Δt,whereN→ total number of photons and λ→wavelength of photons (A) Fx=ΔPxΔt=IAcosθ.sinθC Fy=ΔPyΔt=IAcosθcosθC pr=FyA=Icos2θC (B) Fx=0 Fy=2IAcos2θC pr=2IAcos2θC(C) and (D) Fx=IAcos2θC(1−r) Fy=IAcos2θC(1+r) prIcos2θC(1+r)

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A parallel beam of light is incident on a fixed surface PQ at an angle ' $θ$ ' with the vertical as shown in the figure. The intensity of light is I and area of surface PQ is A. In List-I nature of reflection are given and in List-II force and generated radiation pressure are given. Then match List-I with suitable option List-II. (Given that I = 500 W $/ m^{2}$ , $A = 1.5 m^{2}, θ = 37^{0}$ )
LIST-I LIST-II
P) If all the incident energy is absorbed by surface PQ then force on the surface and radiation pressure generated are respectively. 1) $3.2 \times 10^{- 6} N, 2.13 \times 10^{- 6} N / m^{2}$
Q) If all the incident energy is reflected by surface PQ then force on the surface and radiation pressure generated are respectively. 2) $2.24 \times 10^{- 6} N, 1.38 \times 10^{- 6} N / m^{2}$
R) If 30% incident energy is reflected by the surface then force on the surface and pressure generated are respectively. (Rest absorbed) 3) $2.74 \times 10^{- 6} N, 1.81 \times 10^{- 6} N / m^{2}$
S) If 70% incident energy is reflected back then force on the surface and radiation pressure are respectively. (Rest absorbed) 4) $2 \times 10^{- 6} N, 1.06 \times 10^{- 6} N / m^{2}$
5) $4.2 \times 10^{- 6} N, 3.13 \times 10^{- 6} N / m^{2}$

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$P \to 4; Q \to 1; R \to 3; S \to 2$

$P \to 3; Q \to 2; R \to 1; S \to 4$

$P \to 3; Q \to 5; R \to 4; S \to 2$

$P \to 4; Q \to 1; R \to 2; S \to 3$

answer is D.

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Detailed Solution

Intensity is given by $I = \frac{N (\frac{h C}{λ})}{(A \cos θ) Δ t}$ ,where
$N \to$ total number of photons and $λ \to$
wavelength of photons

$(A) F_{x} = \frac{Δ P_{x}}{Δ t} = \frac{I A \cos θ . \sin θ}{C} F_{y} = \frac{Δ P_{y}}{Δ t} = \frac{I A \cos θ \cos θ}{C} p_{r} = \frac{F_{y}}{A} = \frac{I \cos^{2} θ}{C} (B) F_{x} = 0 F_{y} = \frac{2 I A \cos^{2} θ}{C} p_{r} = \frac{2 I A \cos^{2} θ}{C}$

(C) and (D)
$F_{x} = \frac{I A \cos^{2} θ}{C} (1 - r) F_{y} = \frac{I A \cos^{2} θ}{C} (1 + r) p_{r} \frac{I \cos^{2} θ}{C} (1 + r)$

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