Science, asked by nandnigupta4luck, 3 months ago

Q33. An object of height 5 cm is held 30 cm away from a converging lens of focal length
10 cm. Find the position, size and nature of the image formed. 3​

Answers

Answered by BrainlyTwinklingstar
76

Given :

In convex lens,

Object height = 5cm.

Object distance = - 30cm

Focal length = 10cm.

To find :

The position, size and nature of the image formed.

Solution :

Using lens formula,

The formula which gives the relationship between image distance, object distance and focal length of a lens is known as the lens formula.

The lens formula can be written as :

\boxed{ \bf \dfrac{1}{v} - \dfrac{1}{u}= \dfrac{1}{f}}

where,

  • v denotes image distance
  • u denotes object distance
  • f denotes focal length

by substituting all the given values in the formula,

\leadsto{ \sf \dfrac{1}{v} - \dfrac{1}{u}= \dfrac{1}{f}}

\leadsto{ \sf \dfrac{1}{v} - \dfrac{1}{( - 30)}= \dfrac{1}{10}}

\leadsto{ \sf \dfrac{1}{v}  +  \dfrac{1}{30}= \dfrac{1}{10}}

\leadsto{ \sf \dfrac{1}{v}  =  \dfrac{1}{10} - \dfrac{1}{30}}

\leadsto{ \sf \dfrac{1}{v}  =  \dfrac{3 - 1}{30} }

\leadsto{ \sf \dfrac{1}{v}  =  \dfrac{1}{15} }

\leadsto{ \sf v = 15 \: cm }

thus, the position of image is 15cm.

Now, to find size of image we have to use magnification formula that is

\boxed{ \bf m = \dfrac{h'}{h} =  \dfrac{v}{u}  }

where,

  • m denotes magnification
  • h' denotes image height
  • h denotes object height

by substituting all the given values in the formula

\leadsto{ \sf  \dfrac{h'}{h}  =  \dfrac{v}{u} }

\leadsto{ \sf  \dfrac{h'}{5}  =  \dfrac{15}{ - 30} }

\leadsto{ \sf  {h'}  =  -  \dfrac{15 \times 5}{30} }

\leadsto{ \sf  {h'}  =  -  \dfrac{75}{30} }

\leadsto{ \sf  {h'}  =  2.5 \: cm}

thus, the size of image is 2.5cm

Nature of image

  • The image is real and inverted.
  • The image is diminished.


chocolatelover01: ur answer is very helpful thank you soo much
BrainlyTwinklingstar: welcome
Answered by kandpalsonia85
34

To find position and height of image .

1/f = 1/v - 1/u

1/10 = 1/v - 1/(-30)

1/10 = 1/v + 1/30

1/v = 1/10 - 1/30

1/v = 3-1/30

1/v = 2/30

1/v = 1/15 [ cross multiplication ]

v = 15

ho/hi = v/u

ho/5 = 15/-30

ho = 15×5/30

ho = 2.5cm

nature of image is = inverted ,large in size , real

I hope you will understand !!

please like .

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