Question

list four characteristics of the image formed by a convex lens of focal length 20cm when the object is placed in front of it at a distance of 10 cm from the optical centre​

Answer 1

$\bold{CONVEX \: LENS}$

A convex lens is thick in its middle and thin at the periphery. A light beam converges passing through such a lens is called converging lens.

Given,

◼Object distance v = 10cm

◼Focal length f = 20cm

◼Image distance u = ?

By lens Formula we know,

$\huge \boxed{ \frac{1}{v} - \frac{1}{u}= \frac{1}{f}}$

$\huge \boxed{ \frac{1}{v} = \frac{1}{f} + \frac{1}{u}}$

Since, its convex lens the object distance will be negative

$\frac{1}{v} = \frac{1}{20} + \frac{1}{(-10)}$

$\frac{1}{v} = \frac{1}{20} - \frac{1}{10}$

On solving ,

$\frac{1}{v} = \frac{1 - 2}{20}$

$\frac{1}{v} = \frac{-1}{20}$

So,

$\huge \boxed{v = -20 cm}$

So the object distance is -20 cm.

For linear magnification,

$\boxed{LINEAR \:MAGNIFICATION = \frac{lenght \: of \: image (I) }{lenght \: of \:object(O)}}$

$\boxed{LINEAR \:MAGNIFICATION = \frac{v }{u}}$

so,

$LINEAR \:MAGNIFICATION = \frac{20 }{10}$

$LINEAR \:MAGNIFICATION = \frac{2\cancel{0 }}{1\cancel{0 }}$

$LINEAR \:MAGNIFICATION = 2$

$\bold{CHARACTERISTICS \: OF \: IMAGE}$

▶Since, the object distance is negative therefore, the image is formed on the same side of the object.

▶The image formed will be virtual as its formed at sam size of object.

▶Since, the magnification is more than one therefore the image is magnified.

▶Convex lens can only form such image when kept between focus and optical center, so the image formed will be between second focus and first focus.

Answer 2

Step-by-step explanation:

A convex lens is thick in its middle and thin at the periphery. A light beam converges passing through such a lens is called converging lens.

Given,

◼Object distance v = 10cm

◼Focal length f = 20cm

◼Image distance u = ?

By lens Formula we know,

\huge \boxed{ \frac{1}{v} - \frac{1}{u}= \frac{1}{f}}

v

1

−

u

1

=

f

1

\huge \boxed{ \frac{1}{v} = \frac{1}{f} + \frac{1}{u}}

v

1

=

f

1

+

u

1

Since, its convex lens the object distance will be negative

\frac{1}{v} = \frac{1}{20} + \frac{1}{(-10)}

v

1

=

20

1

+

(−10)

1

\frac{1}{v} = \frac{1}{20} - \frac{1}{10}

v

1

=

20

1

−

10

1

On solving ,

\frac{1}{v} = \frac{1 - 2}{20}

v

1

=

20

1−2

\frac{1}{v} = \frac{-1}{20}

v

1

=

20

−1

So,

\huge \boxed{v = -20 cm}

v=−20cm

So the object distance is -20 cm.

For linear magnification,

\boxed{LINEAR \:MAGNIFICATION = \frac{lenght \: of \: image (I) }{lenght \: of \:object(O)}}

LINEARMAGNIFICATION=

lenghtofobject(O)

lenghtofimage(I)

\boxed{LINEAR \:MAGNIFICATION = \frac{v }{u}}

LINEARMAGNIFICATION=

u

v

so,

LINEAR \:MAGNIFICATION = \frac{20 }{10}LINEARMAGNIFICATION=

10

20

LINEAR \:MAGNIFICATION = \frac{2\cancel{0 }}{1\cancel{0 }}LINEARMAGNIFICATION=

1

0

2

0

LINEAR \:MAGNIFICATION = 2LINEARMAGNIFICATION=2

\bold{CHARACTERISTICS \: OF \: IMAGE}CHARACTERISTICSOFIMAGE

▶Since, the object distance is negative therefore, the image is formed on the same side of the object.

▶The image formed will be virtual as its formed at sam size of object.

▶Since, the magnification is more than one therefore the image is magnified.

▶Convex lens can only form such image when kept between focus and optical center, so the image formed will be between second focus and first focus.