# A concave spherical mirror has a radius of curvature of R = 10.0 cm. a. Calculate the location of...

## Question:

A concave spherical mirror has a radius of curvature of R = 10.0 cm.

a. Calculate the location of image formed by an 14.0 mm tall object whose distance from mirror is 20.0 cm.

b. Calculate the size of image formed by an 14.0 mm tall object whose distance from mirror is 20.0 cm.

c. Calculate the location of image formed by an 14.0 mm tall object whose distance from mirror is 10.0 cm.

d. Calculate the size of image formed by an 14.0 mm tall object whose distance from mirror is 10.0 cm.

e. Calculate the location of image formed by an 14.0 mm tall object whose distance from mirror is 2.50 cm.

f. Calculate the size of image formed by an 14.0 mm tall object whose distance from mirror is 2.50 cm.

g. Calculate the location of image formed by an 14.0 mm tall object whose distance from mirror is 10.0 m.

h. Calculate the size of image formed by an 14.0 mm tall object whose distance from mirror is 10.0 m.

## Concave Spherical Mirrors

A concave spherical mirror is a portion of spherical surface with reflecting properties on the inside. Incident light is focused on a single point and, depending on the position of the reflected object with respect to this focal point, the type of image produced is different.

The distance of the produced image {eq}d_i {/eq} from the focal point is related to the distance {eq}d_o {/eq} of the latter from the object by the mirror equation

{eq}\dfrac{1}{d_i} + \dfrac{1}{d_o} = \dfrac{1}{f} {/eq}

where f is the focal length of the mirror and the distances are defined as positive when the object or the image are in front of the mirror. In the case of a concave spherical mirror, the focal length is simply half of the curvature radius

{eq}f = R / 2 \ . {/eq}

In our case f = 5 cm . Furthermore, the size {eq}h_i {/eq} of the image generated is related to the size {eq}h_o {/eq} of the generating object and their respective distances from the focal point by the so-called magnification equation

{eq}M = \dfrac{h_i}{h_o} = \dfrac{d_i}{d_o} {/eq}

M being defined as the magnification of the mirror. With these two equations, we can easily answer the questions (we will take the "distance from the mirror" to stand for {eq}d_o {/eq} ):

(a) {eq}\dfrac{1}{d_i} + \dfrac{1}{ 20 \ cm} = \dfrac{1}{5 \ cm} \ \longrightarrow \ d_i = 6.67 cm {/eq}

(b) {eq}\dfrac{h_i}{1.40 \ cm} = \dfrac{6.67 \ cm}{20 \ cm} \longrightarrow \ h_i = 4.67 mm {/eq}

(c) {eq}\dfrac{1}{d_i} + \dfrac{1}{ 10 \ cm} = \dfrac{1}{5 \ cm} \ \longrightarrow \ d_i = 10.0 cm {/eq}

(d) {eq}\dfrac{h_i}{1.40 \ cm} = \dfrac{10 \ cm}{10 \ cm} \longrightarrow \ h_i = 14.0 mm {/eq}

(e) {eq}\dfrac{1}{d_i} + \dfrac{1}{ 2.5 \ cm} = \dfrac{1}{5 \ cm} \ \longrightarrow \ d_i = - 5.00 cm {/eq}

(f) {eq}\dfrac{h_i}{1.40 \ cm} = -\dfrac{5 \ cm}{2.5 \ cm} \longrightarrow \ h_i = - 28.0 mm {/eq}

(g) {eq}\dfrac{1}{d_i} + \dfrac{1}{ 1000 \ cm} = \dfrac{1}{5 \ cm} \ \longrightarrow \ d_i = 5.03 cm {/eq}

(h) {eq}\dfrac{h_i}{1.40 \ cm} = \dfrac{5.03 \ cm}{1000 \ cm} \longrightarrow \ h_i = 0.0704 mm \ . {/eq} 