Question

concave maren 0 - 60cm u = 6ocm,f=20cm I = ?. with diagram​

Answer 1

Provided that:

• Concave mirror
• Object distance = 60 cm
• Focal length = 20 cm

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

According to sign convention,

• Concave mirror
• Object distance = -60 cm
• Focal length = -20 cm

To calculate:

• Image distance

Solution:

• Image distance = -30 cm

Using concept:

• Mirror formula

Using formula:

• Mirror formula:

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

Where, 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!

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

LCM of 30 & 60:

$\small{\begin{array}{c|c} \tt  \: \: 2 & \sf{60 - 30} \\ \\ \tt \: \:  2 & \sf {30 - 15} \\ \\ \tt 3 & \sf{15 - 15} \\ \\ \tt  \: \: 5 & \sf{5 - 5} \\ \\ \tt  \: \: 1 & \sf{1 - 1}\\ \\ \tt & \sf{1 - 1} \end{array}}$