Concave Mirror Numerical Class-12 Nootan ISC Physics Solutions Ch-14 Reflection of Light : Spherical Mirrors. Step by step solutions of Kumar and Mittal Physics of Nageen Prakashan as council latest prescribe guideline for upcoming exam. Visit official Website CISCE for detail information about ISC Board Class-12 Physics.
Concave Mirror Numerical Class-12 Nootan ISC Physics Solutions Ch-14 Reflection of Light : Spherical Mirrors
| Board | ISC |
| Class | 12 |
| Subject | Physics |
| Book | Nootan |
| Chapter-14 | Reflection of Light : Spherical Mirrors |
| Topics | Numericals on Concave Mirror |
| Academic Session | 2025-2026 |
Numericals on Concave Mirror
Class-12 Nootan ISC Physics Solutions Ch-14 Reflection of Light : Spherical Mirrors.
Que-1: An object is placed in front of a concave mirror of radius of curvature 40 cm at a distance of 10 cm. Find the position, nature and magnification of the image.
Ans- f = -40/2 = -20 cm , u = -10 cm
now, m = f / f-u
= -20 / -20+10 = 2
again m = -v/u
=> 2 = -v/-10
=> v = 20 cm
Image is twice of object. virtual and 20 cm behind the mirror.
Que-2: A concave mirror has a radius of curvature 30 cm. How far must it be placed from a small object in order to give a five times magnified virtual image?
Ans- f = -30/2 = -15 , m = +5
now, m = f / f-u
=> 5 = -15 / -15-u
=> -15 = -75-u
=> u = -12 cm
-sign shows that object should be placed in front of mirror.
Que-3: A concave mirror produces a real image of half the size of an object placed at 60 cm in front of it. Where should the object be placed to obtain a virtual image of double the size of the object?
Ans- m = -1/2 , u = -60
=> m = f / f-u
=> -1/2 = f / f+60
=> -2f = f+60
=> f = -20 cm
now m = +2 = -20 / -20-u
=> -10 = -20-u
=> u = -10 cm
Que-4: A 1 cm long flame is placed at a distance of 1.5 m from a wall. How far from the wall should a concave mirror be placed in order to obtain a 2 cm long flame on the same wall? What is the focal length of the mirror?
Ans- m = -v/u = -2
=> v = 2u
now v = -x
u = -x/2 = -(x – 1.5)
=> x/2 = x – 1.5
=> x = 3 m
∴u = -1.5 , ∴v = -3 m
again m = f / f-u
=> -2 = f / f+1.5
=> -2f -3 = f
=> f = -1 m
Que-5: An object is placed first at a distance of 25 cm and then at a distance of 40 cm from a concave mirror. The magnification in the first case is four times than that in the second case, the image being real in both cases. What is the focal length of the mirror?
Ans- 4 m2 = m1
4 x f / f+40 = f / f+25
=> 4f + 100 = f + 40
=> 3f = -60
=> f = -20 cm
Que-6: A concave mirror for face viewing has focal length 36.0 cm. Find the distance at which you should hold the mirror from your face in order to see your image upright with a magnification of 4.
Ans- f = -36.0 cm , m = +4
now, m = -v/u
=> 4 = -v/u
=> v = -4u
now using mirror equation
1/f = 1/u + 1/v
=> 1/-36 = 1/u + 1/-4u
=> 4u = -108
=> u = -27 cm
Que-7: An object is placed at a distance of 30 cm from a concave mirror of focal length 15 cm. If the object is displaced through a distance of 10 cm away from the mirror, find the displacement of the image.
Ans- f = -15 cm
u1 = -30 cm
u2 = -40 cm
Using mirror equation
=> 1/v1 = 1/f – 1/u1
=> 1/v1 = 1/-15 – 1/-30
=> v1 = -30 cm
now u2 = -40 cm
=> 1/v2 = 1/-15 – 1/-40
=> v2 = -24 cm
Displacement = Iv2 – v1I
=> I-24 -(-30)I
=> 6 cm
— : End of Concave Mirror Numerical Class-12 Nootan ISC Physics Solutions Ch-14 Reflection of Light : Spherical Mirrors. :–
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