1 Answer
Solution :-
Height of the Object, h0 = 5 cm
Distance of the object from converging lens, u = -25 cm
Focal length of converging lens, f = 10 cm
Using lens formula,
\( \frac{1}{v}-\frac{1}{u}=\frac{1}{f}\)
\( \frac{1}{v}=\frac{1}{f}+\frac{1}{u}=\frac{1}{10}-\frac{1}{25}=\frac{15}{250}\)
\( v = \frac{250}{15}= 16.06 cm\)
Also, for a converging lens, \(\frac{h_i}{h_o}=\frac{v}{u}\)
\( h_i=\frac{v}{u}\times\,h_o=\frac{50\times5}{3\times(-25)}=\frac{10}{-3}= -3.3 cm\)
Thus, the image is inverted and formed at a distance of 16.7 cm behind the lens and measures 3.3 cm. The ray diagram is shown below.
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