Question

an object is 6 cm is infront of convex lens at a distance of 20cm and if it's focal length 25 cm find the position nature and height of image ​

Answer 1

AnsWer :

V = - 100 Cm. and height of image is 30 Cm.

Given :

• u = 20 cm.
• f = 25 cm.
• ho = 6 cm.

To Find :

Position nature and height of image.

Formula :

For Lens,

$\tt \dagger \: \: \: \: \: \dfrac{1}{f} = \dfrac{1}{v} - \dfrac{1}{u}$

Magnified,

$\tt \dagger \: \: \: \: \: m = \dfrac{ v}{u} = \dfrac{ h_i} {h_o}$

For Mirror,

$\tt \dagger \: \: \: \: \: \dfrac{1}{f} = \dfrac{1}{v} + \dfrac{1}{u}$

Magnified,

$\tt \dagger \: \: \: \: \: m = \dfrac{ -v}{u} = \dfrac{ h_i} {h_o}$

Solution :

Let,

• Object distance ( u )
• image distance ( v )
• Focal length ( f )
• height of image ( hi )
• height of object ( ho )

Sign Convention,

• u = - 20 cm.
• f = + 25 cm.

$\rule{90}1$

Now, We have Lens Formula.

$\tt \longmapsto \dfrac{1}{f} = \dfrac{1}{v} - \dfrac{1}{u}$

$\tt\longmapsto \dfrac{1}{25} = \dfrac{1}{v} - \dfrac{1}{ - 20}$

$\tt\longmapsto \dfrac{1}{25} = \dfrac{1}{v} + \dfrac{1}{ 20}$

$\tt\longmapsto \dfrac{1}{25} - \dfrac{1}{ 20} = \dfrac{1}{v}$

$\tt \longmapsto \dfrac{4 - 5}{100} = \dfrac{1}{v}$

$\tt \longmapsto \dfrac{ - 1}{100} = \dfrac{1}{v}$

$\tt\longmapsto 100= - v.$

$\tt \longmapsto v = - 100 \: cm.$

$\rule{90}1$

Now, We have Magnified.

$\tt m = \dfrac{ v}{u} = \dfrac{ h_i} {h_o}$

$\tt\longmapsto m = \dfrac{ - 100}{ - 20}$

$\tt\longmapsto m = \dfrac{ \cancel{ - }10 \cancel{0}}{ \cancel{ -} 2 \cancel{0}}$

$\tt\longmapsto m = \dfrac{ 10}{2}$

$\tt\longmapsto m = \dfrac{ \cancel{10}}{ \cancel2}$

$\tt\longmapsto m = 5.$

$\rule{90}1$

Now,

$\tt \longmapsto m = \dfrac{ h_i} {h_o}$

$\tt \longmapsto 5 = \dfrac{ h_i} {6}$

$\tt \longmapsto h_i = 30 \: cm.$

Therefore,the image formed by lens Virtual erect and height of image is 30 cm.

$\rule{200}3$

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