In a pinhole camera, the image of height 6 cm is formed at a distance of 10 cm. If the object is situated at a distance of 5 cm, the height of the object is ___ cm.

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Solution

Magnification can be related to object distance u and image distance v
Magnification, $m = \frac{v}{u} = \frac{10}{5} = 2$
Or, $\frac{H e i g h t o f i m a g e}{H e i g h t o f o b j e c t} = 2$
Hence, height of object $= \frac{6}{2} = 3 c m$

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In a pinhole camera, the image of height 6 cm is formed at a distance of 10 cm. If the object is situated at a distance of 5 cm, the height of the object is ___ cm.

Magnification can be related to object distance u and image distance v Magnification, m=vu=105=2 Or, Height of imageHeight of object=2 Hence, height of object =62=3 cm

Magnification can be related to object distance u and image distance v
Magnification, $m = \frac{v}{u} = \frac{10}{5} = 2$
Or, $\frac{H e i g h t o f i m a g e}{H e i g h t o f o b j e c t} = 2$
Hence, height of object $= \frac{6}{2} = 3 c m$