Question

an object is placed 15 cm from a ) a converging mirror b) a diverging mirror,of radius of curvature 20 cm .calculate the image position and magnification ..and also draw. Ray diagram of each case.

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Answer 1

Answer:

• Radius of curvature = 20 cm
• Focal length = 10 cm
• Object distance = -15 cm (Object distance is always negative)

CONVERGING MIRROR/CONCAVE MIRROR

• Object distance = -15 cm
• Focal length = -10 cm

1/f = 1/v + 1/u

1/v = 1/f - 1/u

1/v = 1/-10 - 1/-15

1/v = -1/10 + 1/15

1/v = (-15+10)/150

1/v = -5/150

1/v = -1/30

v = -30 cm

Magnification = -v/u

= -(-30)/-15

= 30/-15

= -2

DIVERGING MIRROR/CONVEX MIRROR

• Object distance = -15 cm
• Focal length = +10 cm

1/f = 1/v + 1/u

1/v = 1/f - 1/u

1/v = 1/10 - 1/-15

1/v = 1/10 + 1/15

1/v = 25/150

v = 150/25

v = 6 cm

Magnification = -v/u

= -6/-15

= 6/15

= 0.4

FORMULAS –

• MIRROR FORMULA :– 1/f = 1/v + 1/u
• MAGNIFICATION :– h'/h = -v/u

Answer 2

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QUESTION -

an object is placed 15 cm from:

a ) a converging mirror

b) a diverging mirror,

of radius of curvature 20 cm

Calculate the image position and magnification .

Also draw the Ray diagram of each case.

SOLUTION -

[ 1 ]

• Object Distance , S = -15 cm.

• Focal length = - 10 cm { As Radius Of Curvature = -20 cm }

We know that :

$\dfrac{ 1} { s } + \dfrac{ 1 } { s' } = \dfrac{ 1}{ f }$

Substuting the required values :

$\dfrac{ 1} { -15 } + \dfrac{1}{s'} = \dfrac{1}{-10} \\ \\ \\ => \dfrac{1}{ s'} = \dfrac{ 1 }{15} - \dfrac{ 1}{ 10 } = \dfrac{ -1}{30} \\ \\ \\ => s' = 30 cm$

Magnification :

$Magnification \: = \: \dfrac{ - s' }{ s } = \dfrac{ -30} { -15 } = 2$

So , the required Magnification is 2.

[ 2 ]

Object Distance , S = -15 cm.

Object Distance , S = -15 cm.Focal length = 10 cm { As Radius Of Curvature = 20 cm }

We know that :

$\dfrac{ 1} { s } + \dfrac{ 1 } { s' } = \dfrac{ 1}{ f }$

Substuting the required values :

$\dfrac{ 1} { -15 } + \dfrac{1}{s'} = \dfrac{1}{10} \\ \\ \\ => \dfrac{1}{ s'} = \dfrac{ 1 }{15} + \dfrac{ 1}{ 10 } = \dfrac{ 5}{30} \\ \\ \\ => s' = 6 cm$

Magnification :

$Magnification \: = \: \dfrac{ - s' }{ s } = \dfrac{ -6} { -15 } = 0.4$

So , the required Magnification is 0.4.

Ray Diagrams -

[ 1 ] Refer to attachment 1.

[ 2 ] Rfefer to attachment 2.