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CBSE Class 10 Science 2016 Term II Delhi Set 2

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Question : 11 of 11
Marks: +1, -0
(a) Draw a ray diagram to show the formation of image by a concave lens when an object is placed in front of it.
(b) In the above diagram mark the object distance (u)(u) and the image-distance (v)(v) with their proper signs (+ve or -ve as per the new Cartesian sign convention) and state how these distances are related to the focal length (f)(f) of the concave lens in this case.
(c) Find the nature and power of a lens which forms a real and inverted image of magnification -1 at a distance of 40 cm from its optical centre.
Solution:  
(b) The object distance (u)(u) and image distance (v)(v) are marked in the diagram of part (a).
Relation :   1f=  1v−  1u\;\frac{1}{f}=\;\frac{1}{v}-\;\frac{1}{u}
(c) As, m=−1m=-1; hence, the lens is convex
Now, m=  vum=\;\frac{v}{u}
∴    v=−u\therefore \;\; v=-u
Thus, object is at 2f2 f.
∵    v  =40 cm   (Given)   \because \;\; v\;=40 \text{ cm} \;\text{ (Given) }\;
∴    2f  =40 cm\therefore \;\; 2 f\;=40 \text{ cm}
f  =20 cm=0.2 m.f\;=20 \text{ cm}=0.2 \text{ m} .
P=  1f  =  10.2=+5 D   (convex lens)   P=\;\frac{1}{f}\;=\;\frac{1}{0.2}=+5 \text{ D} \;\text{ (convex lens) }\;
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