Question

1) An object with a height of 36 cm is placed 2.1 m in front of a concave mirror with a focal length of 0.50 m. a) Determine the location and size of the image using a ray diagram. b) Is the image upright or inverted? c) Is the image real or virtual?

2) Using the mirror equations, find the:
a. Precise location of the image.
b. Magnification of the image.
c. Height of the image.

Answer 1

The image will be inverted because the value of the magnification is negative. The image formed will be virtual as the object is placed in front of the focal length.

(a) Location of image v=0.65 m

(b) Magnification of the object m=-1.3

(c)Height of the image will be =46.8 cm

What is a concave mirror?

The concave mirror has the reflective surface inwards and focuses the light to the focal point. The image formed in the concave mirror can be virtual or real if the object is placed in front of the concave mirror image formed will be virtual.

The formula for the concave mirror is given as:

$\dfrac{1}{v} =\dfrac{1}{u} -\dfrac{1}{f}$

here

v= Distance of image

u= distance of object

f=focal length

It is given that

u=2.1 m

f=0.50 m

h= 36 cm

(a) Now the location of the image:

$\dfrac{1}{v} =\dfrac{1}{u} -\dfrac{1}{f}$

$\dfrac{1}{v} =\dfrac{1}{2.1} -\dfrac{1}{0.5}$

$v=-0.65 \ m$

(b) Magnification of the image:

$M=\dfrac{v}{u}$

$M=\dfrac{-0.65}{0.5}$

$M=-1.3$

(c) Height of the image:

$M=\dfrac{H_i}{H_o}$

$1.3=\dfrac{H_i}{36}$

$H_i=46\ cm$

Hence image will be inverted because the value of the magnification is negative. The image formed will be virtual as the object is placed in front of the focal length.

(a) Location of image v=0.65 m

(b) Magnification of the object m=-1.3

(c)Height of the image will be =46.8 cm

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