Question

A concave lens of focal length 15 cm forms an image at a distance of 10 cm from the lens. How far is the object from the lens? What is its nature and magnification?​

Answer 1

Answer:

here f=15cm nd v= -10 cm..

thus by formula-

solve..

1/f = 1/v - 1/u

understand?

Answer 2

Given:

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• f = – 15 cm
• v = – 10 cm

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where ‘f’ means focal length of the concave lens and ‘v’ means distance of the image from the lens.

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To find:

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Distance of the object from the lens.

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Solution:

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We know that focal length of the concave lens is always negative so ‘f’ is negative here and ‘v’ i.e. distance of the image from the lens will also be negative because v is always –ive for a concave lens.

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Now by applying lens formula i.e.

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

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$\begin{gathered}\begin{gathered} \\ \\ : \implies \displaystyle \frac{1}{ - 10} - \frac{1}{u} = \frac{1}{ - 15} \\ \\ \\ \end{gathered} \end{gathered} \\ \\ \begin{gathered}\begin{gathered} \\ \\ : \implies \displaystyle \frac{1}{u} = \frac{1}{ - 10} + \frac{1}{15} \\ \\ \\ = \frac{ - 3 + 2}{30} = \frac{1}{ - 30} \\ \\ \\ \end{gathered} \end{gathered}$

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So, u = – 30 cm.

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It means the object is at a distance of 30 cm from the concave lens.

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We know that,

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magnification = distance of the image/distance of the object from the lens.

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so, m = v/u

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$= \frac{ - 10 \: cm}{ - 30 \: cm} \: \\ \\ = \frac{1}{3} = \: + \: 0.33$

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Here magnification is positive. So, the image is erect, virtual and diminished by 0.33 times.

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Useful formula:

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Lens formula → 1/v – 1/u = 1/f

Linear magnification → m = ${h_i}$ / ${h_o}$

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Note:

1. In case of a convex lens, the magnification “m” is positive, when image is virtual, and negative when image is real.
2. In case of a concave lens, the magnification ‘m’ is always positive.

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