Question

If an object of 10 cm height is placed at distance of 30 cm from concave mirror of focal length 12 cm ,find the position nature of image​

Answer 1

Provided that:

• Concave mirror is given.
• Height of object = 10 cm.
• Distance of object = 30 cm.
• Focal length = 12 cm

* Don't use the above information in your answer as here we haven't use sign convention till now!

According to sign convention,

• Height of object = +10 cm
• Distance of object = -30 cm
• Focal length = -12 cm

To determine:

• Position of image
• Nature of image

Concepts to be imply:

• 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!

$:\implies \sf \dfrac{1}{v} + \dfrac{1}{u} = \dfrac{1}{f} \\ \\ :\implies \sf \dfrac{1}{v} + \dfrac{1}{-30} = \dfrac{1}{-12} \\ \\ :\implies \sf \dfrac{1}{v} - \dfrac{1}{30} = - \dfrac{1}{12} \\ \\ :\implies \sf \dfrac{1}{v} = - \dfrac{1}{12} + \dfrac{1}{30} \\ \\ :\implies \sf \dfrac{1}{v} = \dfrac{5 \times -1 + 2 \times 1}{60} \\ \\ :\implies \sf \dfrac{1}{v} = \dfrac{-5 + 2}{60} \\ \\ :\implies \sf \dfrac{1}{v} = \dfrac{-3}{60} \\ \\ :\implies \sf \dfrac{1}{v} = \dfrac{-1}{20} \\ \\ :\implies \sf 1 \times 20 = v \times -1 \\ \\ :\implies \sf 20 = -v \\ \\ :\implies \sf -20 = v \\ \\ :\implies \sf v \: = -20 \: cm \\ \\ :\implies \sf Image \: distance \: = -20 \: cm \\ \\ {\pmb{\sf{Henceforth, \: solved!}}}$

~ Now calculating magnification!

$:\implies \sf m = \dfrac{h'}{h} = \dfrac{-v}{u} \\ \\ :\implies \sf m = \dfrac{-v}{u} \\ \\ :\implies \sf m = \dfrac{-(-20)}{-30} \\ \\ :\implies \sf m = \dfrac{20}{-30} \\ \\ :\implies \sf m = \dfrac{-2}{3} \\ \\ :\implies \sf m = -0.6 \\ \\ :\implies \sf Magnification = -0.6$

~ Calculating height of image!

$:\implies \sf m = \dfrac{h'}{h} = \dfrac{-v}{u} \\ \\ :\implies \sf m = \dfrac{h'}{h} \\ \\ :\implies \sf -0.6 = \dfrac{h'}{10} \\ \\ :\implies \sf -0.6 \times 10 \: = h' \\ \\ :\implies \sf -6 \: = h' \\ \\ :\implies \sf Image \: height \: = -6 \: cm$

Henceforth,

• Distance of image / Position of image = -20 centimetres from mirror.

• Magnification = -0.6

• Height of image / size of image = -6 centimetres.

• Image is real and inverted.

• Image is enlarged.

____________________

LCM of 12 & 30:

$\small{\begin{array}{c|c} \tt  \: \: 2 & \sf{12 - 30} \\ \\ \tt \: \:  2 & \sf {6 - 15} \\ \\ \tt 3 & \sf{3 - 15} \\ \\ \tt  \: \: 5 & \sf{1 - 5} \\ \\ \tt  \: \: 1 & \sf{1 - 1}\end{array}}$