an object is placed at i)10cm, ii)5cm in front of a concave mirror of radius of curvature 15 cm. find the position, nature, and magnifiaction of the image in each case.
Answers
1) Object distance = 10 cm
Object distance = u = -10 cm
Radius of curvature = R = -15 cm
Focal length = f
f = 2R
so,
f = R/2 = -15/2 cm
We know that :-
1/v + 1/u = 1/f
v = image distance
1/v = 1/f - 1/u
= 1 / (-15/2) - (-1/10)
= -2/15 + 1/10
= -4/30 + 3/30
= -1/30
1/v = -1/30
v = -30 cm
Image distance = v = -30 cm
The position of image is 30 cm in front of mirror.
we know that ,
magnification , m = -v/u
= -(-30)/-10
= -3
Nature :- This negative sign indicates that the image is real ,and inverted.
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ii) Object distance = 5 cm
Object distance = u = -5cm
Focal length = -15/2 cm
1/v = 1/f - 1/u
= -2/15 - (-1/5)
= -2/15 + 1/5
= 1/15
v = 15 cm
Image distance = v = 15 cm
we know ,
m = -v/u
= -15/-5
= 3
Nature of image :- The positive sign of magnification indicates that the image is virtual , erect and magnified
Answer:
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Explanation:
) Object distance = 10 cm
Object distance = u = -10 cm
Radius of curvature = R = -15 cm
Focal length = f
f = 2R
so,
f = R/2 = -15/2 cm
We know that :-
1/v + 1/u = 1/f
v = image distance
1/v = 1/f - 1/u
= 1 / (-15/2) - (-1/10)
= -2/15 + 1/10
= -4/30 + 3/30
= -1/30
1/v = -1/30
v = -30 cm
Image distance = v = -30 cm
The position of image is 30 cm in front of mirror.
we know that ,
magnification , m = -v/u
= -(-30)/-10
= -3
Nature :- This negative sign indicates that the image is real ,and inverted.
--------------------------------------------------------------------
ii) Object distance = 5 cm
Object distance = u = -5cm
Focal length = -15/2 cm
1/v = 1/f - 1/u
= -2/15 - (-1/5)
= -2/15 + 1/5
= 1/15
v = 15 cm
Image distance = v = 15 cm
we know ,
m = -v/u
= -15/-5
= 3
Nature of image :- The positive sign of magnification indicates that the image is virtual , erect and magnified