PLEASE EXPLAIN AND SHOW WORK. Thank You. Answer is already shown. An object plac

Question

PLEASE EXPLAIN AND SHOW WORK. Thank You. Answer is already shown.

An object placed in front of a concave mirror, with radius of curvature of 30 cm, forms an image 25 cm behind the lens with a height of 8 cm. 5.4 cm 6.4 cm 7.4 cm 8.4 cm 9.4 cm What is the hight of the object? 1 cm 2 cm 3 cm 4 cm 5 cm Radio waves (f = 107.1 MHz, with k = 2.8 m) diffract through an opening with width of 3.8 m. At what angle above the central maximum will no signal reach (where should you not place your radio)? 27 degree 37 degree 47 degree 57 degree 67 degree If you tune into a lower frequency station (f = 100.3 MHz, with lambda = 3.0 m) what will happen to this angle? it will increase it will decrease it will stay the same

Explanation / Answer

For a concave mirror
1/f = 1/v + 1/u
here v = -25 (image formed behind the lens ); radius of curvature = 30 cm ; f = 15 cm
1/15 = -1/25 + 1/u
1/v = 1/15 + 1/25
1/v = 5+3/75 = 8/75
v = 75/8cm = 9.375 cm
Therefore the object is kept 9.375 cm in front of the mirror.
now magnification (m) = -v/u
also m = Hi/Ho where Hi is Image height and Ho is object height
therefore

Hi = 8 cm
m = -1*-25/9.375 = 8/Ho
Ho = 8*9.375 / 25 = 3cm

3) for the radio wave to show no signal at an angle t (theta )

k = wavelength = 2.8 m

d = slit width = 3.8 m

therefore , sint = 2.8/3.8 = .7368

t = sin-1(.7368) = .8284 rad = 47.49 o

4 )

now if k = 3m

then , sint = 3/3.8 = .7895

t = sin-1(.7895) = .90995 rad = 52.163o

therefore , the angle will increase

dsint = k

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