Question

An object is placed 15 cm from a convex mirror of radius of curvature 60cm .What will be the position of image from the mirror?

1 point

A. 4.0cm

B. 7.5 cm

C.7.0 cm

D.6.0 cm

​

Answer 1

Provided that:

• Distance of object = 15 cm
• Radius of curvature = 60 cm

* Don't use the above information in your solution as it let your answer to be wrong as here we haven't use sign convention till now.

According to sign convention,

• Distance of object = -15 cm
• Radius of curvature = + 60 cm

* Use that information in your answer where you have applied sign convection.

To determine:

• Position of image

Solution:

• Position of image = 10 cm

Using concepts:

• Mirror formula
• Focal length formula

Using formulas:

• Mirror formula:

• ${\small{\underline{\boxed{\pmb{\sf{\dfrac{1}{v} + \dfrac{1}{u} = \dfrac{1}{f}}}}}}}$

• Focal length formula:

• ${\small{\underline{\boxed{\pmb{\sf{\dfrac{R}{2} \: = f}}}}}}$

Where, R denotes radius of curvature , v denotes image distance, f denotes focal length and u denotes object distance.

Knowledge required:

• If the focal length is positive then the mirror is “Convex Mirror.”

• If the focal length is negative then the mirror is “Concave Mirror.”

• Object distance is always negative in both the mirrors that are concave mirror and convex mirror.

Required solution:

~ Firstly let us find out the focal length by using suitable formula!

$:\implies \sf \dfrac{R}{2} \: = f \\ \\ :\implies \sf \dfrac{60}{2} \: = f \\ \\ :\implies \sf 30 = f \\ \\ :\implies \sf f \: = 30 \: cm \\ \\ :\implies \sf 30 = f \\ \\ :\implies \sf Focal \: length \: = 30 \: cm \\ \\ {\pmb{\sf{Henceforth, \: solved!}}}$

~ Firstly let us find out the position of the image by using mirror formula!

$:\implies \sf \dfrac{1}{v} + \dfrac{1}{u} = \dfrac{1}{f} \\ \\ :\implies \sf \dfrac{1}{v} + \dfrac{1}{-15} = \dfrac{1}{30} \\ \\ :\implies \sf \dfrac{1}{v} - \dfrac{1}{15} = \dfrac{1}{30} \\ \\ :\implies \sf \dfrac{1}{v} = \dfrac{1}{30} + \dfrac{1}{15} \\ \\ :\implies \sf \dfrac{1}{v} = \dfrac{1 \times 1 + 2 \times 1}{30} \\ \\ :\implies \sf \dfrac{1}{v} = \dfrac{1 + 2}{30} \\ \\ :\implies \sf \dfrac{1}{v} = \dfrac{3}{30} \\ \\ :\implies \sf \dfrac{1}{v} = \dfrac{1}{10} \\ \\ :\implies \sf 1 \times 10 = v \times 1 \\ \\ :\implies \sf 10 = v \\ \\ :\implies \sf v \: = 10 \: cm \\ \\ :\implies \sf Image \: distance \: = 10 \: cm \\ \\ {\pmb{\sf{Henceforth, \: solved!}}}$

Position of image = 10 centimetres, not 4.0 cm , 7.5 cm , 7.0 cm or 6.0 cm.

Answer 2

Answer:

Provided that:

Distance of object = 15 cm

Radius of curvature = 60 cm

* Don't use the above information in your solution as it let your answer to be wrong as here we haven't use sign convention till now.

According to sign convention,

Distance of object = -15 cm

Radius of curvature = + 60 cm

* Use that information in your answer where you have applied sign convection.

To determine:

Position of image

Solution:

Position of image = 10 cm

Using concepts:

Mirror formula

Focal length formula

Using formulas:

• Mirror formula:

{\small{\underline{\boxed{\pmb{\sf{\dfrac{1}{v} + \dfrac{1}{u} = \dfrac{1}{f}}}}}}}

v

1

+

u

1

=

f

1

v

1

+

u

1

=

f

1

• Focal length formula:

{\small{\underline{\boxed{\pmb{\sf{\dfrac{R}{2} \: = f}}}}}}

2

R

=f

2

R

=f

Where, R denotes radius of curvature , v denotes image distance, f denotes focal length and u denotes object distance.

Knowledge required:

• If the focal length is positive then the mirror is “Convex Mirror.”

• If the focal length is negative then the mirror is “Concave Mirror.”

• Object distance is always negative in both the mirrors that are concave mirror and convex mirror.

Required solution:

~ Firstly let us find out the focal length by using suitable formula!

\begin{gathered}:\implies \sf \dfrac{R}{2} \: = f \\ \\ :\implies \sf \dfrac{60}{2} \: = f \\ \\ :\implies \sf 30 = f \\ \\ :\implies \sf f \: = 30 \: cm \\ \\ :\implies \sf 30 = f \\ \\ :\implies \sf Focal \: length \: = 30 \: cm \\ \\ {\pmb{\sf{Henceforth, \: solved!}}}\end{gathered}

:⟹

2

R

=f

:⟹

2

60

=f

:⟹30=f

:⟹f=30cm

:⟹30=f

:⟹Focallength=30cm

Henceforth,solved!

Henceforth,solved!

~ Firstly let us find out the position of the image by using mirror formula!

\begin{gathered}:\implies \sf \dfrac{1}{v} + \dfrac{1}{u} = \dfrac{1}{f} \\ \\ :\implies \sf \dfrac{1}{v} + \dfrac{1}{-15} = \dfrac{1}{30} \\ \\ :\implies \sf \dfrac{1}{v} - \dfrac{1}{15} = \dfrac{1}{30} \\ \\ :\implies \sf \dfrac{1}{v} = \dfrac{1}{30} + \dfrac{1}{15} \\ \\ :\implies \sf \dfrac{1}{v} = \dfrac{1 \times 1 + 2 \times 1}{30} \\ \\ :\implies \sf \dfrac{1}{v} = \dfrac{1 + 2}{30} \\ \\ :\implies \sf \dfrac{1}{v} = \dfrac{3}{30} \\ \\ :\implies \sf \dfrac{1}{v} = \dfrac{1}{10} \\ \\ :\implies \sf 1 \times 10 = v \times 1 \\ \\ :\implies \sf 10 = v \\ \\ :\implies \sf v \: = 10 \: cm \\ \\ :\implies \sf Image \: distance \: = 10 \: cm \\ \\ {\pmb{\sf{Henceforth, \: solved!}}}\end{gathered}

:⟹

v

1

+

u

1

=

f

1

:⟹

v

1

+

−15

1

=

30

1

:⟹

v

1

−

15

1

=

30

1

:⟹

v

1

=

30

1

+

15

1

:⟹

v

1

=

30

1×1+2×1

:⟹

v

1

=

30

1+2

:⟹

v

1

=

30

3

:⟹

v

1

=

10

1

:⟹1×10=v×1

:⟹10=v

:⟹v=10cm

:⟹Imagedistance=10cm

Henceforth,solved!

Henceforth,solved!

Position of image = 10 centimetres, not 4.0 cm , 7.5 cm , 7.0 cm or 6.0 cm.