A concave mirror with a radius of curvature of 20 cm forms an image of a 5.0 cm-

Question

A concave mirror with a radius of curvature of 20 cm forms an image of a 5.0 cm-tall object. The image is located 15 cm in front of the mirror. a) Sketch the principle-ray diagram. b) The object is located ___________________________________________. c) The height of the image is ______________. d) The image is (a) real or (b) virtual and is (c) erect or (d) inverted.

Two charged plates separated by 2.00cm creates a uniform electric field of strength 104 N/C. A charged particle with a mass of 6.64x10-27 kg is released from rest and travels 2.00cm between the two plates in 2.84x10-7s. Show that the charge of the particle is 3.29 x 10-19C.

Explanation / Answer

1)

given
R = 20 cm
h = 5.0 cm
v = 15 cm

so, f = R/2 = 20/2 = 10 cm
a)

b) let u is the object distance

use, 1/u + 1/v = 1/f

1/u = 1/f - 1/v

1/u = 1/10 - 1/15

u = 30 cm

c) magnification, m = -v/u

= -15/30

= -0.5

object height, h' = m*h

= -0.5*5

= -2.5 cm (negative sign indicates the image is inverted)

d) The image is real and inverted

2)

given

s = 2.00 cm = 2*10^-2 m
m = 6.64*10^-27 kg
E =10^4 N/c
t = 2.84*10^-7 s

let a is the acceleration of the charged particle.

force acting on the charged particle,

F = q*E

m*a = q*E

a = q*E/m

now use, s = u*t + (1/2)*a*t^2

s = 0 + (1/2)*a*t^2 (since initial velocity, u = 0 )

2*s/t^2 = a

2*s/t^2 = q*E/m

==> q = 2*s*m/(E*t^2)

= 2*2*10^-2*6.64*10^-27/(10^4*(2.84*10^-7)^2)

= 3.293*10^-19 C

= 3.29*10^-19 C (on rounding to three significant figures) <<<<<<<----Answer

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