2. C+ -3 points SerPSE9 36.P013 An object of height 2.70 cm is placed 33.0 cm fr

Question

2. C+ -3 points SerPSE9 36.P013 An object of height 2.70 cm is placed 33.0 cm from a convex spherical mirror of focal length of magnitude 12.5 cm. (a) Find the location of the image. se a negative value if the image is behind the mirror. (b) Indicate whether the image is upright or inverted. O upright O inverted (c) Determine the height of the image. Need Help? Read it -3 points SerPSE9 36.P025.MI A spherical mirror is to be used to form an image 5.90 times the size of an object on a screen located 6.00 m from the object. (a) Is the mirror required concave or convex? What is the required radius of curvature of the mirror? b) (c) Where should the mirror be positioned relative to the object? m from the object Need Help? Read It l Master It -2 points SerPSE9 36.PO41.W The projection lens in a certain slide projector is a single thin lens. A slide 24.1 mm high is to be projected so that its image fills a screen 1.71 m high The slide-to-screen distance is 2.99 m. (a) Determine the focal length of the projection lens b) How far from the slide should the lens of the projector be placed to form the image on the screen?

Explanation / Answer

2. Given that :

object height, h0 = 2.7 cm

object distance, d0 = 33 cm

focal length of convex spherical mirror, f = - 12.5 cm

(a) the locations of the image is given as ::

using a mirror formula,

1 / d0 + 1 / di = 1 / f

1 / di = 1 / f - 1 / d0

1 / di = 1 / (-12.5 cm) - 1 / (2.7 cm)

1 / di = (-15.2 / 33.75) cm

di = -(33.75 / 15.2) cm

di = - 2.22 cm                      (image is in -ve that means image behind the mirror)

(b) the image is inverted.

(c) height of the image, hi is given as ::

magnification, m = - di / d0

(hi / h0) = - di / d0

then, hi / (2.7 cm) = - (-2.22 cm) / (33 cm)

hi = (2.7 cm) (2.22 cm) / (33 cm)

hi = 0.18 cm

4. Given that :

object height, h0 = 24.1 mm

image height, hi = 1.71 m = 1710 mm

slide-To-screen distance, d0 + di = 2.99 m = 2990 mm

(a) the focal length of the projection lens will be given as ::

using magnification formula, m = di / d0

(hi / h0) = di / d0

di / d0 = (1710 mm) / (24.1 mm)

di / d0 = 70.9 mm

di = (70.9 mm) d0

Now, we have d0 + di = (2990 mm)

then, d0 + (70.9 mm) d0 = (2990 mm)

d0 = (2990 mm) / (71.9 mm)

d0 = 41.5 mm

and   di = (70.9 mm) d0 = (70.9 mm) (41.5 mm)

di = 2942.3 mm

Using a lens formula,    1 / d0 + 1 / di = 1 / f

1 / (41.5 mm) + 1 / (2942.3 mm) = 1 / f

1 / f = (2983.8 / 122105.45) mm

f = (122105.45 / 2983.8) mm

f = 40.92 mm

(b) distance from slides to screen which is given as :

d0 = (2990 mm) - di

d0 = (2990 mm) - (2942.3 mm)

d0 = 47.7 mm

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