PART A: An 18-cm-long pencil is placed beside a convex spherical mirror and its

Question

PART A:

An 18-cm-long pencil is placed beside a convex spherical mirror and its image is 11.5 cm in length. If the radius of curvature of the mirror is 89.6 cm, find the image distance, the object distance, and the magnification of the pencil.

image distance______ cm

object distance______ cm

magnification

PART B:

A shiny sphere, 25 cm in diameter, is placed in a garden for aesthetic purposes. Determine the type, location, and height of the image of a 6.00-cm-tall squirrel that is located 35 cm in front of the sphere.

dI= _____cm

hI= ______cm

Explanation / Answer

A)
Object height, ho = 18 cm
Image height, hi = 11.5 cm
Take u and v as the object distance and the image distance.
focal length = R/2
= 89.6/2
= 44.8 cm

ho/hi = (18/11.5) = 1.565 = - u/v
1/v = - 1.565/u ....(1)

Using the equation 1/f = 1/u + 1/v
f is negative for convex mirrors, substituting (1)
1/(-f) = 1/u + [-1.565/u]
-1/44.8 = - 0.565/u
u = 44.8 x 0.565
= 25.32 cm

From (1), v = - u/1.565
= - 25.32/1.565
= - 16.18 cm

Magnification = hi/ho
= 11.5/18
= 0.64

Image distance, v = -16.18 cm
Object distance, u = 25.32 cm
Magnification, m = 0.64

B)
Radius = 25/2 = 12.5 cm
Focal length, f = R/2 = 6.25 cm
Object distance, u = 35 cm
Take v as the image distance

Using the equation
1/f = 1/u + 1/v
1/v = 1/(-f) - 1/u
= - 1/6.25 - 1/35
v = - 5.3 cm

Magnification = - v/u = hi/ho
5.3/35 = hi/6
hi = 0.91 cm

Image is virtual, upright, located at 5.3 cm and its height is 0.91 cm

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