Question

An object is placed in front of a convex lens of a length 10cm. What is the nature of the image formed if the object distance is 15 cm?​

Answer 1

Explanation:

When an object is placed 10 cm in front of lens A, the image is real, inverted, magnified and formed at a great distance. When the same is placed 10 cm in front of lens B, formed is real, inverted and same size as the object.

Answer 2

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${ \bold { \underline{Question:-}}}$

• An object is placed in front of a convex lens of a length 10cm. What is the nature of the image formed if the object distance is 15 cm?

$\text{\large\underline{\red{Given:-}}}$

• Focal length of lens = 10cm
• Distance of object = 15cm
• Type of lens : concex

$\text{\large\underline{\orange{To Find:-}}}$

• Distance of image and nature of image.

$\text{\large\underline{\purple{Solution:-}}}$

Focal length of convex lens is taken positive and that of concave lens is taken negative.

Distance of image can be calculated by using lens formula..

• $\sf \: \dfrac{1}{u} + \dfrac{1}{v} = \dfrac{1}{f}$

• $\sf \: \dfrac{1}{v} - \dfrac{1}{ - 15} = \dfrac{1}{10}$

• $\sf \: \dfrac{1}{v} + \dfrac{1}{15} = \dfrac{1}{10}$

• $\sf \: \dfrac{1}{v} = \dfrac{1}{10} - \dfrac{1}{15}$

• $\sf \: \dfrac{1}{v} = \dfrac{1}{30}$

• v = 30 cm

In order to find nature of image, we need to calculate magnification first.

• m = $\sf \: \dfrac{v}{u}$

• m = $\dfrac{– 30}{– 15}$

• m = - 2

${ \bold { \underline{Nature\:of\:Image :-}}}$

• Real Inverted
• Enlarged

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