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Concave Mirror Reflection Problems

1. A point object is placed between a plane mirror and concave mirror facing each other. The distance between the mirrors is 22.5 cm and the concave mirror has a radius of curvature of 20 cm. For the final image to form on the object itself after two reflections, the object must be placed 15 cm from the concave mirror. 2. A point object oscillating with an amplitude of 2 mm is placed 15 cm from a concave mirror of radius 20 cm. The amplitude of the image formed will be 8 mm.

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78 views1 page

Concave Mirror Reflection Problems

1. A point object is placed between a plane mirror and concave mirror facing each other. The distance between the mirrors is 22.5 cm and the concave mirror has a radius of curvature of 20 cm. For the final image to form on the object itself after two reflections, the object must be placed 15 cm from the concave mirror. 2. A point object oscillating with an amplitude of 2 mm is placed 15 cm from a concave mirror of radius 20 cm. The amplitude of the image formed will be 8 mm.

Uploaded by

tarak das
Copyright
© © All Rights Reserved
We take content rights seriously. If you suspect this is your content, claim it here.
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Download as PDF, TXT or read online on Scribd
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CPP-1 Class - XI Batches - PHONON

REFLECTION

1. A point object is kept between a plane mirror and a concave mirror facing each other. The distance between the mirrors
is 22.5 cm. The radius of curvature of the concave mirror is 20 cm. What should be the distance of the object from the
concave mirror so that after two successive reflections the final image is formed on the object itself ?
[Consider first reflection from concave mirror] :
(A) 5 cm (B*) 15 cm (C) 10 cm (D) 7.5 cm
Sol. Distance between object and cancave mirror
First reflection from concave mirror : 22.5 cm R = 20 cm
1 1 1 f = 10 cm
 = O
v1 ( x) (10)
x
1 1 1

 v ( x) = 
1 10
Plane Concave
1 1 1 10  x
 v =  =
1 x 10 10 x
10 x
v1 =
10  x
 it is negative
Image of concave mirror becomes object for plane mirror. Second reflection from plane mirror image is formed at same
distance from plane as is the distance of object. Now according to question, second image is formed at object itself.
So :
  10 x 
   10  x  – 22.5 = 22.5 – x
  
10 x
 x– = 22.5 + 22.5 = 45
(10  x )
10 x  x 2  10 x
 = 45
10  x
 –x2 = 45 (10 – x)
2
 x – 45x + 450 = 0

45  452  4(450) 45  2025  1800


x= =
2 2
45  225 45  15 60 30
= = = or
2 2 2 2
x = 30 or 15
x = 30 is not possible, so :
x = 15 Ans.

2. A point object at 15 cm from a concave mirror of radius of curvature 20 cm is made to oscillate along the principal axis
with amplitude 2 mm. The amplitude of its image will be :
(A) 2 mm (B) 4 mm (C*) 8 mm (D) 16 mm
v
Sol. Transverse magnification = mT = 
u

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