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A 5 cm long needle is placed along the principal axis of a concave mirror of a focal length 10 cm. It is observed that one end of the image of the needle coincides with one of the ends of the needle. The other end of the image is at a distance x from the pole of the mirror, where x is:-  (More than 1 answer is possible)

A. 20

B. 50/3

C. 30

D. 10

2 Answers

+1 vote
by (367 points)
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Best answer

CORRECT OPTION -: B AND C

GIVEN, 

1 END OF NEEDLE IS COINCIDING WITH THE ANOTHER END OF IMAGE SO IT MUST BE AT CENTER OF CURVATURE (20 CM). AS THE NEEDLE IS 5 CM LONG SO IT CAN EITHER BE AT 

CASE 1) 5 CM AWAY FROM MIRROR

HENCE 

U= -25 CM

F = -10 CM

1/F=1/V+1/U

-1/10+1/25=1/V

V= -50/3

CASE 2) NEEDLES 2ND POINT IS TOWARDS MIRROR

HENCE, 

U= -15

F= -10

1/F=1/V+1/U

-1/10+1/15=1/V

V= -30 CM

Thanks ♥ 

–1 vote
by (40.5k points)
edited by

Correct option is (A) 20 and (B) 50/3

Given f = 10 cm

Let R = 2f

v = -2f

v = -2 x 10

VA = -20 cm

For B u = -25

f = -10

\(\frac1v=\frac1f-\frac1u\) 

\(\frac1v=\frac1{-10}+\frac1{25}\) 

\(\frac1v=\frac{-25+10}{250}\) 

\(\frac1v=\frac{-15}{250}=\frac{-3}{50}\)

\(\frac1v=-\frac3{50}\)

VB = -50/3

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