Question

the focal length of a converging lens is 20cm . an object is 60cm from the lens . where will the image is formed and what type of image is it

Answer 1

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FOCAL LENGTH OF A CONVEX LENS :

=> 20CM

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THE DISTANCE FROM THE OBJECT ;

=> U :

=> -60 CM

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ACTUALLY THE DISTANCE GIVEN IS 60 CM

=> ACCORDING TO SIGN CONVENSIONS ,

=> THE OBJECT IS PLACED AT THE LEFT SIDE OF THE LENS

=> AND SO, THE MEASUREMENT IS TAKEN AS NEGATIVE

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TO FIND :

=> THE DISTANCE FROM LENS TO IMAGE :

=> THAT IS 'V'

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FOCAL LENGTH :

=> $\frac{1}{f} = \frac{1}{v} - \frac{1}{u}$

=> $\frac{1}{20} = \frac{1}{v} - \frac{1}{ - 60}$


=> $\frac{1}{v} = \frac{1}{20} + \frac{1}{60}$


=> $\frac{1}{v} = \frac{3}{60} + \frac{1}{60}$

=> $\frac{1}{v} = \frac{3 + 1}{60}$

=> $\frac{1}{v} = \frac{4}{60}$

=> $\frac{1}{v} = \frac{1}{15}$

=> $v = 15$

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SO ,

=> IMAGE IS FORMED AT A DISTANCE OF 15 CM FROM THE LENS

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=> THUS IMAGE IS REAL AND INVERTED

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THANKS .....

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Answer 2

Answer:

Given:

Object Distance (u) = - 60cm

Focal Length (f₁) = 20cm

To find: Image Distance (v)

From the Lens formula,

1/f = 1/v - 1/u

We get, v = f x u/ f + u

= v = 20 x (-60)/ 20 - 60

= 20 x 60/40 = 30cm

⇒ Image Distance = 30cm

∴ The image is formed in between F₁ and 2F₁.

  Nature of the Image is (a) Real (b) Dimenished (c) Inverted

(ii) Given:

Height of the Object () = 2cm

To find: Height of the Image ()

We know that,

=  =  ⇒  = 30/-60 = 1/-2 ⇒  ⇒  = -1cm

∴ Height of the Image is 1cm, which is below the principal axis.

⇒ For more Clarification, you can observe the figure in the Attachment.