Physics, asked by OmRangari, 10 months ago

 A concave lens has focal length of 20 cm. At what distance from the lens a 5 cm tall object be placed so that it forms an image at 15 cm from the lens? Also calculate the size of the image formed.​

Answers

Answered by Anonymous
77

Explanation:

According to the question,

==> Image distance, v = -15cm

(negative due to convex lens)

==> Focal length, f = -20cm

Let the object distance be u

By lens formula ,

 \frac{1}{v}  -  \frac{1}{u}  =  \frac{1}{f}  \\  \\  \frac{1}{ - 15}  -  \frac{1}{u}  =  \frac{1}{ - 20}  \\  \\  \frac{1}{u}  =  \frac{1}{20}  +  \frac{1}{ - 15}  \\  \\  \frac{1}{u}  =  \frac{3 - 4}{60}  =  \frac{ - 1}{60}  \\  \\

==> u = -60 cm

Therefore, object is placed at 60 cm away from the lens, on the same side as image.

Now ,

Height of object , h1 = 5cm

magnification , \: m \:  =  \frac{h2}{h1}  =  \frac{v}{u}

Putting value of v and u ,

Magnification,

m \:  =  \frac{h2}{5}  =  \frac{ - 15}{ - 60}  \\  \\ ==> \frac{h2}{5}  =  \frac{1}{4}  \\  \\ ==> \: h2 \:  =  \frac{5}{4}  = 1.25cm

Thus , the height of the image is 1.25cm and the positive sign means the image is virtual and erect .

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