Question

A10 mm long awl pin is placed vertically in front of a concave mirror. A 5 mm long image of the awl pin is formed at 30 cm in front of the mirror. The focal length of this mirror is
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Answer 1

➭ Questíon :-

$⟶$A10 mm long awl pin is placed vertically in front of a concave mirror. A 5 mm long image of the awl pin is formed at 30 cm in front of the mirror. The focal length of this mirror is

➭ Gíven :-

$⟶$Size of an awl pin, h = 10 mm

$⟶$Size of the image formed of an awl pin, h' = 5 mm

Image distance from the mirror of an awl $⟶$pin, v = -30 cm (in front of the mirror)

➭ To fínd :-

$⟶$The focal length of the mirror.

➭ Formula requíred :-

$⟶ \frac{h'}{h} = \frac{ - v}{u} \: \: \: \: \: \: \\ ⟶ \frac{1}{f} = \frac{1}{v} + \frac{1}{u}$

➭ Solutíon :-

We can find the object distance using the formula of magnification such that

$\boxed{\sf \frac{h'}{h} = \frac{ - v}{u}}$

u is the object distance and v is image distance

$⟹ u = \frac{ - vh}{h'} \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \\ ⟹ u = \frac{ - 30cm \times 10cm}{5cm} \\ ⟹ u = - 60cm \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \:$

Let f is the focal length of the mirror. Using mirror's formula we get :-

$⟹ \frac{1}{f } = \frac{1}{v} + \frac{1}{u} \: \: \: \: \: \: \: \: \: \: \: \\ ⟹ \frac{1}{f} = \frac{1}{ - 30} + \frac{1}{ - 60} \\ ⟹ f = - 20cm \: \: \: \: \: \: \: \: \: \: \:$

hence, the focal length of the mirror is 20 cm.

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☞ More to know.....

• The focal length ( f ) is the distance from a lens or mirror to the focal point ( F ).
• This is the distance from a lens or mirror at which parallel light rays will meet.
• Light rays that are parallel to each other, but not to the optical axis, will meet on the focal plane.