Question

An object, 6 cm in size is placed 2 points at 30 cm in front of a concave mirror of focal length 20 cm. At what distance from the mirror the image will form? Find nature and size of the image.​

Answer 1

Provided that:

• Height of object = 6 cm
• Distance of object = 30 cm
• Focal length = 20 cm

* Don't use the above information in your solution as it let your answer to be wrong as here we haven't use sign convention till now.

According to sign convention,

• Height of object = + 6 cm
• Object distance = -30 cm
• Focal length = -20 cm

* Use the information when you have applied sign convention.

To determine:

• Distance of image
• Nature of image
• Size of image

Using concepts:

• Magnification formula mirror.
• Mirror formula.

Using formulas:

• Magnification formula:

• ${\small{\underline{\boxed{\pmb{\sf{m = \dfrac{h'}{h} = \dfrac{-v}{u}}}}}}}$

• Mirror formula:

• ${\small{\underline{\boxed{\pmb{\sf{\dfrac{1}{v} + \dfrac{1}{u} = \dfrac{1}{f}}}}}}}$

Where, m denotes magnification, h′ denotes height of the image, h denotes height of the object, v denotes image distance, f denotes focal length and u denotes object distance.

Knowledge required:

• If the magnification produced by a spherical mirror is in negative then the mirror is always “Concave Mirror.”

• If the magnification produced by a spherical mirror is in positive then the mirror is always “Convex Mirror.”

• If magnification is negative in a concave mirror then it's nature is “Real and Inverted” always.

• If magnification is positive then it's nature is “Virtual and Erect” always.

• If in the ± magnification, magnitude > 1 then the image formed is “Enlarged”.

• If in the ± magnification, magnitude < 1 then the image formed is “Diminished”.

• If in the ± magnification, magnitude = 1 then the image formed is “Same sized”.

• If the focal length is positive then the mirror is “Convex Mirror.”

• If the focal length is negative then the mirror is “Concave Mirror.”

Required solution:

~ Firstly let us find out the distance of the image by using mirror formula!

$:\implies \sf \dfrac{1}{v} + \dfrac{1}{u} = \dfrac{1}{f} \\ \\ :\implies \sf \dfrac{1}{v} + \dfrac{1}{-30} = \dfrac{1}{-20} \\ \\ :\implies \sf \dfrac{1}{v} - \dfrac{1}{30} = - \dfrac{1}{20} \\ \\ :\implies \sf \dfrac{1}{v} = - \dfrac{1}{20} + \dfrac{1}{30} \\ \\ :\implies \sf \dfrac{1}{v} = \dfrac{3 \times -1 + 2 \times 1}{60} \\ \\ :\implies \sf \dfrac{1}{v} = \dfrac{-3 + 2}{60} \\ \\ :\implies \sf \dfrac{1}{v} = \dfrac{-1}{60} \\ \\ :\implies \sf 1 \times 60 = v \times -1 \\ \\ :\implies \sf 60 = -v \\ \\ :\implies \sf -60 = v \\ \\ :\implies \sf v \: = -60 \: cm \\ \\ :\implies \sf Image \: distance \: = -60 \: cm \\ \\ {\pmb{\sf{Henceforth, \: solved!}}}$

~ Now let us find out the magnification of the image produced by the mirror!

$:\implies \sf m = \dfrac{-v}{u} \\ \\ :\implies \sf m = \dfrac{-(-60)}{-30} \\ \\ :\implies \sf m = \dfrac{+60}{-30} \\ \\ :\implies \sf m = \dfrac{-6}{3} \\ \\ :\implies \sf m = - 2 \: cm \\ \\ :\implies \sf Magnification = - 2 \: cm \\ \\ {\pmb{\sf{Henceforth, \: solved!}}}$

~ Now let us find out the height of the image!

$:\implies \sf m = \dfrac{h'}{h} \\ \\ :\implies \sf -2 = \dfrac{h'}{6} \\ \\ :\implies \sf -2 \times 6 \: = h' \\ \\ :\implies \sf h' \: = -12 \: cm \\ \\ {\pmb{\sf{Henceforth, \: solved!}}}$

Therefore,

• Distance of image from pole of the mirror = -60 centimetres.

• Height of the formed image is -12 centimetres.

• The image is real and inverted.

• The image is enlarged.

___________________

LCM of 20 and 30:

$\large{\begin{array}{c|c} \tt  \: \: 2 & \sf{30 - 20} \\ \\ \tt \: \:  2 & \sf {15 - 10} \\ \\ \tt 3 & \sf{15 - 5} \\ \\ \tt  5 & \sf{5 - 5} \\ \\ \tt  \: \: 1 & \sf{1 - 1}\\ \\ \tt & \sf{1 - 1} \end{array}}$