Question

) A student wants to project the image of a candle flame on a screen 100 cm in front of a mirror by keeping the candle flame at a distance of 25 cm from its pole. (i) Which type of mirror should he use? (ii) Find the magnification of the image produced. (iii) Find the focal length of the mirror​

Answer 1

Provided that:

→ Image of a candle flame is on a screen 100 cm in front of a mirror.

→ Candle flame is at a distance of 25 cm from the pole of mirror.

According to sign convention:

• Distance of image = -100 cm
• Distance of object = -25 cm

To determine:

• Which type of mirror is it!?
• Magnification!?
• Focal length!?

Solutions:

• It is concave mirror.
• Magnification = -4
• Focal length = -20 cm

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 in a concave mirror 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 focal length of the given mirror by using mirror formula!

$:\implies \sf \dfrac{1}{v} + \dfrac{1}{u} = \dfrac{1}{f} \\ \\ :\implies \sf \dfrac{1}{-100} + \dfrac{1}{-25} = \dfrac{1}{f} \\ \\ :\implies \sf - \dfrac{1}{100} - \dfrac{1}{25} = \dfrac{1}{f} \\ \\ :\implies \sf \dfrac{1 \times -1 - 4 \times 1}{100} = \dfrac{1}{f} \\ \\ :\implies \sf \dfrac{-1 -4}{100} = \dfrac{1}{f} \\ \\ :\implies \sf \dfrac{-5}{100} = \dfrac{1}{f} \\ \\ :\implies \sf \dfrac{-1}{20} = \dfrac{1}{f} \\ \\ :\implies \sf -1 \times f = 20 \times 1 \\ \\ :\implies \sf -f = 20 \\ \\ :\implies \sf -20 \: = f \\ \\ :\implies \sf f \: = -20 \: cm \\ \\ :\implies \sf Focal \: length \: = -20 \: cm \\ \\ {\pmb{\sf{Henceforth, \: solved!}}}$

~ Now let's determine which mirror is it!?

→ We noticed that focal length is in negative therefore, the mirror is concave mirror.

~ Now let's calculate the magnification of the image produced by using magnification of mirror formula!

$:\implies \sf m = \dfrac{h'}{h} = \dfrac{-v}{u} \\ \\ :\implies \sf m = \dfrac{-v}{u} \\ \\ :\implies \sf m = \dfrac{-(-100)}{-25} \\ \\ :\implies \sf m = \dfrac{+100}{-25} \\ \\ :\implies \sf m = -4 \\ \\ :\implies \sf Magnification = -4 \\ \\ {\pmb{\sf{Henceforth, \: solved!}}}$