Question

A 2.0 cm tall object is placed 35 cm in front of a concave mirror of 21 cm focal length.
(a) Show your calculation of the image location.
(b) Explain why the image is real or virtual.
(c) Show your calculation of the image height.
(d) Explain why the image is upright or inverted.
(e) Show your complete graphical solution to this problem.
(f) On graph paper, place the principle axis, mirror, object, and
focal length to scale. Add three principle rays to the graph to locate and define the image.

Answer 1

As per the given question,

height of the object (hi)=2cm

distance of the object from the mirror (u)=35cm

Focal length of the concave mirror (f)=21cm

let the distance of the image from the mirror is (v)=?

i) We know that,

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

$\Rightarrow \dfrac{1}{v}-\dfrac{1}{35}=\dfrac{1}{-21}$

$\Rightarrow \dfrac{1}{v}=\dfrac{-1}{21}+\dfrac{1}{35}=-\dfrac{1}{7}(\dfrac{1}{3}-\dfrac{1}{5})$

$\Rightarrow \dfrac{1}{v}=-\dfrac{1}{7}(\dfrac{5-3}{15})$

$\Rightarrow v=\dfrac{-105}{2}=-52.5cm$

$ii)m=\dfrac{v}{u}=\dfrac{52.5}{35}=+1.5$

So, Image will be real

iii) Let the Height of the image be hi

$m=\dfrac{-hi}{ho}$

$\Rightarrow hi=-m \times ho$

$\Rightarrow hi=-1.5\times 2 = -3 cm$

iv) From the above, we can see the height of the object is +ive and the height of the image is coming negative, here negative sign representing that is in the negative y axis that means image is inverted.

v)