An object is placed at a distance of 6cm in front of a mirror is found to have an image 15cm (a) in front
of it (b) behind it. Find the focal length and the kind of mirror used in both the cases.
Answers
Explanation:
We know that,
We know that,v
We know that,v1
We know that,v1
We know that,v1 +
We know that,v1 + u
We know that,v1 + u1
We know that,v1 + u1
We know that,v1 + u1 =
We know that,v1 + u1 = f
We know that,v1 + u1 = f1
We know that,v1 + u1 = f1
We know that,v1 + u1 = f1
We know that,v1 + u1 = f1 u = -15 cm
We know that,v1 + u1 = f1 u = -15 cmFocal length = -10 cm
We know that,v1 + u1 = f1 u = -15 cmFocal length = -10 cm∴
We know that,v1 + u1 = f1 u = -15 cmFocal length = -10 cm∴ v
We know that,v1 + u1 = f1 u = -15 cmFocal length = -10 cm∴ v1
We know that,v1 + u1 = f1 u = -15 cmFocal length = -10 cm∴ v1
We know that,v1 + u1 = f1 u = -15 cmFocal length = -10 cm∴ v1 =
We know that,v1 + u1 = f1 u = -15 cmFocal length = -10 cm∴ v1 = −10
We know that,v1 + u1 = f1 u = -15 cmFocal length = -10 cm∴ v1 = −101
We know that,v1 + u1 = f1 u = -15 cmFocal length = -10 cm∴ v1 = −101
We know that,v1 + u1 = f1 u = -15 cmFocal length = -10 cm∴ v1 = −101 −
We know that,v1 + u1 = f1 u = -15 cmFocal length = -10 cm∴ v1 = −101 − (−15)
We know that,v1 + u1 = f1 u = -15 cmFocal length = -10 cm∴ v1 = −101 − (−15)1
We know that,v1 + u1 = f1 u = -15 cmFocal length = -10 cm∴ v1 = −101 − (−15)1
We know that,v1 + u1 = f1 u = -15 cmFocal length = -10 cm∴ v1 = −101 − (−15)1
We know that,v1 + u1 = f1 u = -15 cmFocal length = -10 cm∴ v1 = −101 − (−15)1 ⇒
We know that,v1 + u1 = f1 u = -15 cmFocal length = -10 cm∴ v1 = −101 − (−15)1 ⇒ v
We know that,v1 + u1 = f1 u = -15 cmFocal length = -10 cm∴ v1 = −101 − (−15)1 ⇒ v1
We know that,v1 + u1 = f1 u = -15 cmFocal length = -10 cm∴ v1 = −101 − (−15)1 ⇒ v1
We know that,v1 + u1 = f1 u = -15 cmFocal length = -10 cm∴ v1 = −101 − (−15)1 ⇒ v1 =
We know that,v1 + u1 = f1 u = -15 cmFocal length = -10 cm∴ v1 = −101 − (−15)1 ⇒ v1 = −10
We know that,v1 + u1 = f1 u = -15 cmFocal length = -10 cm∴ v1 = −101 − (−15)1 ⇒ v1 = −101
We know that,v1 + u1 = f1 u = -15 cmFocal length = -10 cm∴ v1 = −101 − (−15)1 ⇒ v1 = −101
We know that,v1 + u1 = f1 u = -15 cmFocal length = -10 cm∴ v1 = −101 − (−15)1 ⇒ v1 = −101 +
We know that,v1 + u1 = f1 u = -15 cmFocal length = -10 cm∴ v1 = −101 − (−15)1 ⇒ v1 = −101 + 15
We know that,v1 + u1 = f1 u = -15 cmFocal length = -10 cm∴ v1 = −101 − (−15)1 ⇒ v1 = −101 + 151
We know that,v1 + u1 = f1 u = -15 cmFocal length = -10 cm∴ v1 = −101 − (−15)1 ⇒ v1 = −101 + 151
We know that,v1 + u1 = f1 u = -15 cmFocal length = -10 cm∴ v1 = −101 − (−15)1 ⇒ v1 = −101 + 151 =
We know that,v1 + u1 = f1 u = -15 cmFocal length = -10 cm∴ v1 = −101 − (−15)1 ⇒ v1 = −101 + 151 = 30
We know that,v1 + u1 = f1 u = -15 cmFocal length = -10 cm∴ v1 = −101 − (−15)1 ⇒ v1 = −101 + 151 = 30−3+2
We know that,v1 + u1 = f1 u = -15 cmFocal length = -10 cm∴ v1 = −101 − (−15)1 ⇒ v1 = −101 + 151 = 30−3+2
We know that,v1 + u1 = f1 u = -15 cmFocal length = -10 cm∴ v1 = −101 − (−15)1 ⇒ v1 = −101 + 151 = 30−3+2 =−
We know that,v1 + u1 = f1 u = -15 cmFocal length = -10 cm∴ v1 = −101 − (−15)1 ⇒ v1 = −101 + 151 = 30−3+2 =− 30
We know that,v1 + u1 = f1 u = -15 cmFocal length = -10 cm∴ v1 = −101 − (−15)1 ⇒ v1 = −101 + 151 = 30−3+2 =− 301
We know that,v1 + u1 = f1 u = -15 cmFocal length = -10 cm∴ v1 = −101 − (−15)1 ⇒ v1 = −101 + 151 = 30−3+2 =− 301
We know that,v1 + u1 = f1 u = -15 cmFocal length = -10 cm∴ v1 = −101 − (−15)1 ⇒ v1 = −101 + 151 = 30−3+2 =− 301
We know that,v1 + u1 = f1 u = -15 cmFocal length = -10 cm∴ v1 = −101 − (−15)1 ⇒ v1 = −101 + 151 = 30−3+2 =− 301 ⇒v=−30cm
We know that,v1 + u1 = f1 u = -15 cmFocal length = -10 cm∴ v1 = −101 − (−15)1 ⇒ v1 = −101 + 151 = 30−3+2 =− 301 ⇒v=−30cmMagnification (m)=
We know that,v1 + u1 = f1 u = -15 cmFocal length = -10 cm∴ v1 = −101 − (−15)1 ⇒ v1 = −101 + 151 = 30−3+2 =− 301 ⇒v=−30cmMagnification (m)= u
We know that,v1 + u1 = f1 u = -15 cmFocal length = -10 cm∴ v1 = −101 − (−15)1 ⇒ v1 = −101 + 151 = 30−3+2 =− 301 ⇒v=−30cmMagnification (m)= uv
We know that,v1 + u1 = f1 u = -15 cmFocal length = -10 cm∴ v1 = −101 − (−15)1 ⇒ v1 = −101 + 151 = 30−3+2 =− 301 ⇒v=−30cmMagnification (m)= uv
We know that,v1 + u1 = f1 u = -15 cmFocal length = -10 cm∴ v1 = −101 − (−15)1 ⇒ v1 = −101 + 151 = 30−3+2 =− 301 ⇒v=−30cmMagnification (m)= uv =
We know that,v1 + u1 = f1 u = -15 cmFocal length = -10 cm∴ v1 = −101 − (−15)1 ⇒ v1 = −101 + 151 = 30−3+2 =− 301 ⇒v=−30cmMagnification (m)= uv = Object height
We know that,v1 + u1 = f1 u = -15 cmFocal length = -10 cm∴ v1 = −101 − (−15)1 ⇒ v1 = −101 + 151 = 30−3+2 =− 301 ⇒v=−30cmMagnification (m)= uv = Object heightImage hheig
Also, Iage height=−2×6=−12 cm
Also, Iage height=−2×6=−12 cmTherefore, a real, inverted and enlarged image of the object is formed beyond the centre of curvature.