Physics, asked by Anonymous, 10 months ago

Solve the given Questions Correctly.

1.an object is placed 9 cm from the concave lens of focal length 12 cm find the magnification, position of the image, nature of the image ?

2.an object of size 3 centimetres is placed at a distance of 18 cm away from the convex lens of radius of curvature R 1 is equal to 7 cm,R2= 13 cm then find the magnification, image distance, size of image ,position of image​

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Answers

Answered by rajsingh24
21

1)ANSWER :-

NATURE of IMAGE IS VIRTUAL, ERECT AND

MAGNIFICATION (M=1.5)

M=P/Q (.°. V/U)

M=4/6

M=-3/2g

M=-1.5

2)ANSWER:-

Nature of Image is Virtual and Erect.

&

Size of the Image: Diminished.

M=V/U

M=3/18

(M=6)

Attachments:
Answered by nirman95
20

Answer:

1st question:

Applying Lens formula :

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

  =  >  \frac{1}{ - 12}  =  \frac{1}{v}  -  (\frac{1}{ - 9})  \\

 =  >  \frac{1}{v}  =  -  \frac{1}{12}  -  \frac{1}{9}  \\

 =  > v =  - \frac{36}{7} cm \\

So , the image will be formed at 36/7 cm.

Since the object is placed before the focus, we will get a virtual, erect and diminished image.

2nd question:

We shall use Lens Maker's Formula :

 \frac{1}{v}  -  \frac{1}{u}  = ( \frac{n2}{n1}  - 1)( \frac{1}{r1}  -  \frac{1}{r2} ) \\

Putting value of u = -18, n2 = 3/2 , n1 = 1 , r1 = +7 cm and r2 = -13 cm, we get:

1/v = 1/(0.0548)

=> v = 18.24 cm.


Anonymous: Fabulous
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