Question

An object is placed in front of half of a concave mirror. (2m)

a. Make a ray diagram. Will an image form at the position indicated? Why or why not?

b. When the object is 20 cm from the mirror, the image is half the height of the object. Determine the focal length of the mirror.

Answer 1

Question 1:

1. An object is placed in front of half of a concave mirror Make a ray diagram. Will an image form at the position indicated? Why or why not?

Solution:

• Image will be formed as usual following the rules of ray diagram formation. The only difference will be in the fact that the brightness (intensity) of the image will be comparitively less.

• Thats because , half of the concave mirror is absent and does not pass on the incident rays.

Question 2:

When the object is 20 cm from the mirror, the image is half the height of the object. Determine the focal length of the mirror.

Solution:

$\rm{ \therefore \: magnification = \dfrac{ h_{i} }{ h_{0} } = - \dfrac{v}{u} }$

$\rm{ = > \: \dfrac{ - \frac{1}{2} h_{0}}{ h_{0} } = - \dfrac{v}{u} }$

$\rm{ = > \: - \dfrac{1}{2} = - \dfrac{v}{u} }$

$\rm{ = > v = \dfrac{u}{2} }$

Applying Mirror Formula:

$\therefore \: \sf{ \dfrac{1}{f} = \dfrac{1}{v} + \dfrac{1}{u} }$

$= > \: \sf{ \dfrac{1}{f} = \dfrac{2}{u} + \dfrac{1}{u} }$

$= > \: \sf{ \dfrac{1}{f} = \dfrac{ 2 + 1}{u} }$

$= > \: \sf{ \dfrac{1}{f} = \dfrac{ 3}{u} }$

$= > \: \sf{ f = \dfrac{ u}{3} }$

$= > \: \sf{ f = \dfrac{ ( - 20)}{3} }$

$= > \: \sf{ f = - 6.67 \: cm}$

So, final answer is:

$\boxed{ \red{\: \sf{ f = - 6.67 \: cm}}}$