Question

draw a ray diagram to show the formation of image of an object placed between infinity and the optical centre of a concave lens concave lens of focal length 15cm forms an image 10 cm from the lens calculate the distance of the object form from the lens the magnification for the image formed with the nature of the image formed​

Answer 1

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Answer 2

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(a) Draw a ray diagram to show the formation of image of an object placed between infinity and optical centre of a concave lens.

(b) A concave lens of focal length 15 cm forms an image 10 cm from the lens. Calculate (i) the object distance, (ii) the nature and magnification of the image formed.

The following Ray diagram illustrates how an object situated between a concave lens' optical centre and infinity forms an image:

Between the optical centre and the focus, the image will be obtained. The image that is created will be virtual, upright, and smaller in size.

In the question, Focal length of concave lens $f = -15 cm$

Image distance $v= -10 cm$

The formula of lens is $\frac{1}{f}=\frac{1}{v}-\frac{1}{u}$ .

We need to get the value of $u$ .

The expression can be written as $\frac{1}{u}=\frac{1}{v}-\frac{1}{f}$ .

Put the values in the expression .

Therefore,the expression is  $\frac{1}{u}=\frac{1}{-10}-\frac{1}{-15}$ .

The value of $\frac{1}{u}=-\frac{1}{30}$ .

The distance from the object is $u=-30cm$ .

The distance of the object form from the lens is $-30 cm$ .

Magnification is determined by the picture size to object size ratio and can be expressed as follows:

The equation of magnification is $m=\frac{v}{u}$ .

Substitute the value to get $m=\frac{-10}{-30}$ .

The value of $m=\frac{1}{3}$ .

The image created will be erect and smaller because the magnification is positive and less than 1.

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