Q.459. Concave Mirror Problem.

Question 459.
Suppose your height is h, and you are standing in front of a concave mirror of focal length f where your real image is height h/8.73. What is your distance from the mirror in terms of f.

Answer 459.
(h/8.73)/h = v/u
=> v/u = 1/(8.73) ... ( 1 )

1/v + 1/u = 1/f
=> v = fu/(u - f)
=> v/u = f/(u - f) ... ( 2 )

From ( 1 ) and ( 2 ),
f/(u - f) = 1/(8.73)
=> u - f = 8.73f
=> u = 9.73f.

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Q.379. Refraction of light.

Question 379.
You need to hit an underwater target lying flat on the bottom of a pool. The water is 1 m deep and you are standing so that your eyes are 3 m above the bottom of the pool. As you look at the target your gaze is 30 degrees below the horizontal. At what angle below the horizontal should you throw a spear to hit the target? (assume it travels in a straight line and that you throw at the same level as your eyes).

Answer 379.
Refer to the figure shown below.

For refraction,
sin i / sin r = 4/3
=> sin r = (3/4) sin(60°) = 3√3/8
=> tan r = tan [arcsin (3√3/8)]
=> tan r = 0.8542421962
=> AB = 0.8542421962

MB = 2cot (60°) = 3.4641016151

=> MA
= MB + BA
= 0.8542421962 + 3.4641016151
= 4.31834381

=> required angle,
θ = arctan (3/4.31834381) = 34.79°.

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Q.207. Total internal reflection, refraction of light, prism.

Question 207.

Refer to the figure as under. Monochromatic light is incident normally on the face OY of a prism made of glass whose index of refraction is such that the critical angle is 42 degrees.

Redraw this figure showing the path of light through the prism, any reflections as well as refractions, until some of the light emerges from the prism.

Answer 207.
As the incident ray is normal to the hypotenuse, it passes undeviated and hits the vertical surface of the prism at 30 degrees with the vertical or 60 degrees with the norm.al at 8 cm below O.

As the critical angle is 42 degrees, the ray wiull undergo total internal reflection as its incident angle is 60 degrees which is more than the critical angle and hit the bottom surface at some point with the incident angle of 30 degrees and come out with refraction.

If r = angle of refraction                       

sin i / sin r = sinC

=> sinr = sin i / sinC = sin30° / sin42°
=> r = 48.35 degrees.

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Q.184. Refraction of light

Question 184.

What is the final y-position of the ray?

Answer 184.

For the green slab, the angle of refraction, r, is given by

sinr / sin35° = 1/1.3

=> r = 26.1813°

=> y-position at interface

= 5 tan(26.1813°) cm

= 2.4583 cm

For the blue slab, the angle of refraction, r', is given by         

sinr' / sinr = 1.3/1.8

=> r' = (1.3)/1.8) * sin(26.1813°)

= 18.5815°
=> final y-position
= 2.4583 cm + 5 tan(18.5815°) cm
= (2.4583 + 1.6809) cm
= 4.1392 cm.

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Q.168. Interpherence, Young's Double-slit Formula

Question 168.
A police cruiser has an unusual radar speed trap set up. It has two transmitting antennae at the edge of a main road that runs north and south. One antenna is 2.0 m[west] of the other, and they are both fed from a common transmitter with a frequency of 3.0 x10^9 Hz. These antennae can be considered as point sources of continuous radio waves. The trap is set for cars travelling south. A student drives his car along an east-west road that crosses the main road at a level intersection 100 m[north] of the radar trap. The student's car has a "radar detector", so he hears a series of beeps as he drives west through the intersection. If the time interval between successive quiet spots is 1/5 s, as he crosses the main road, what is his speed?

Answer 168.
This is the problem of Young's double-slit interference experiment.

d = 2.0 m
D = 100 m
λ = c/f = 3x10^8 / 3 x 10^9 = 10^(-1) m
distance between successive destructive interference,
x = λD/d = 10^(-1) * 100/2 = 5 m
=> speed of the car
= 5/(1/5) m/s
= 25 m/s
= 25 * 3.6 km/h
= 90 km/h.

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Q.156. Refraction of Light

Question 156.
Ryan Baloney stands beside his swimming pool completely full of water. (the pool, not Ryan!) (Refer to Fig.)
His eyes are located 1.6 m directly above the edge as shown, and he looks directly across the 7.00 m wide pool. Light from the bottom corner opposite him arrives at his eye in such a direction that the apparent depth of the water, h, is 0.30 m. If the index of refraction for water is 4/3 calculate H, the actual depth of the pool. Answer in m.

Answer 156.
Refr to the following figure for the additional construction of refracted ray from the bottom corner of the pool.

Using the property of similar triangles,
h / x = 1.6 / (7 - x)
=> (0.30) (7 - x) = 1.6x
=> 1.9 x = 2.1
=> x = (2.1) / (1.9) m

sin i = x / √(x^2 + h^2)

sinr = x / √(x^2 + H^2)

sini / sinr = 4/3 = √(x^2 + H^2) / √(x^2 + h^2)

=> 16 (x^2 + h^2) = 9 (x^2 + H^2)

=> 9H^2 = 7x^2 + 16h^2

=> H

= (1/3) √(7x^2 + 16h^2)

= (1/3) √[7 * (2.1/1.9)^2 + 16 * (0.3)^2]

= (1/3) √(8.5512 + 1.44)

= 1.054 m.

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Q.93. Wave Optics.

Question 93.

The attenuation coefficient of the muscle is 800 cm^-1, at the wavelength emitted by CO2 laser

(1.06 x 10^-5). Calculate the thickness of the muscle layer that absorbs 90% of the light energy of this laser.

Answer 93.

I / I'= e^ ( - μ x )

where
I = intensity of emergent ray
I' = intensity of original ray
μ = attenuation coefficient
x = thickness

As energy is proportional to intensity, putting given information in this formula noting that 90% is absorbed so 10% remains which means I / I' = 0.10

=>0.1 = e ^ ( - 800 x ), where x = thickness in cm.
=> ln (0.1) = - 800x
=> ln 1 - ln 10 = - 800x
=> x = (ln 10) / 800 = 0.003 cm = 0.03 mm.
[Note: wavelength is not required here.]
Source(s):
Refer to page 9 of the .pdf page in the link which gives the formula that can be used to solve your problem.
http://www.physics.uoguelph.ca/~pgarrett/teaching/PHY-1070/lecture-21.pdf#page=9

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