Assignment 4, solutions:

Problem 1:

(a)  Consider some surface S enclosing a point P.  Now imagine a small cone, which intersects an infinitesimal area, dA, on S.  The cone defines the solid angle subtended by the area dA at point P.  By definition, the solid angle is the area dA projected on a plane perpendicular to the radius vector r from P to dA, and divided by r².  If dW is the solid angle subtended by dA we  have 

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The expression can be integrated over a region of S to find the total solid angle subtended by that region

Thus if the apex of a cone lies at the center of a sphere, the solid angle subtended is the ratio of the sphere surface area enclosed by the cone to the square of the radius of the sphere.

If cosq’ = a/(a²+r²)^1/2, then

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Therefore 

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(b)  If a goes to infinity, then a/(a²+r²)^1/2 goes to 1 and W = 2p(1 - 1) = 0.

(c)  If P is very close to the disk, then a is very small compared to R.  In zeroth order approximation we neglect the a/(a²+r²)^1/2 term and W = 2p.  The cone encloses the surface area of half of a sphere.

(d)  If R/a <<1 then we expand

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Therefore W = pR²/a².  The surface area of the sphere enclosed by the cone is so small that we can neglect its curvature and treat it as flat.

Problem 2:

(a)  The illuminance is the photometric equivalent of the irradiance.  It is the amount of light energy received from a source by the surface dS’, measured in lm/m².

The book is far from the lamp, so we can assume that the solid angle an area dA of the book subtends at the lamp is dA/r², where r = 4 m.    The power per unit solid angle emitted by the lamp is I = 100 lm/sr.  Therefore we have for the illuminance 

E = I/r² = (100 lm/sr)(1/r²)sr = (100/16) lm/m² = 6.25 lm/m².

(b)  Light reflected back from the mirror forms a parallel beam.  All light reflected from a projected area of the mirror equal to the area of the book will subsequently fall onto the book.  Since the mirror is 3 m from the book, a projected area of the mirror equal to the area of the book intercepts 16/9 times as much of the light as the book does.  It reflects 90% of that light, so the increase in the illuminance is 

DE = (100/16)(16/9)0.9 lm/m² = 10 lm/m².

Problem 3:

Let N be the number of photons incident per second.  If each photon releases 1 electron, the N is also the number of electrons released per second.

The incident power is f = NE_photon.

The current is given by I = DQ/Dt = Ne.

The responsivity is I/f = e/E_photon = (1.6×10^-19 C)/(2×1.6×10^-19J) = 0.5 A/W.

Problem 4:

(a)  Lens 1 is the aperture stop.  Its image in object space is lens 1 itself.  It is therefore also the entrance pupil.  Its image in image space is located 4f/3 to the right of lens 2.  This is the exit pupil.  The diameter of the exit pupil is D' = |M|D = (1/3)D.

(b)  In object space the image of lens 1 is lens 1 itself.  The tangent half-angle subtended by lens 1 at the on-axis position of the entrance pupil is infinite.  The image of lens 2 is located at 4f/3 to the left of lens 1.  The diameter of the image is D' = |M|(3/2)D = (1/2)D.  The half-angle subtended at the on-axis position of the entrance pupil in the small angle approximation is 3D/16f.   So the image of lens 2 formed by lens 1 is the entrance window.  Lens 2 is the field stop.  The exit window is lens 2 itself.

(c)  The entrance window is located 4f/3 to the left of lens 1.  The diameter of the entrance window is D' = (1/2)D.  The half-angle subtended at the on-axis position of the entrance pupil is 3D/16f.  Twice this angle, 3D/8f, is the field of view.

Problem 5:

(C)  The solid angle subtended by the detector when the source is located at the center of the face is 2p.  So the detector is 100% efficient in detecting gamma rays of the specific energy.

The solid angle subtended by the detector when the detector is moved 1 m away from the source is W = (pd²/4)/R², where d is the diameter of the detector, R is its perpendicular distance from the source, and d << R.  The fraction of gamma rays detected is W/4p.  Here W/4p =(p0.08²/4) = 4*10^-4.

Problem 6:

(D)  The Wien Law gives the wavelength of the peak of the radiation distribution, l_max = 2.9´10^-3 m K/T.  Here l is measured in units of meter and T is in Kelvin.
We have l_max ~ 2´10^-6 m, so T ~ 1.5´10³ K.