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A height of a tower formed by a objective lens of a telescope is 4.7 cm. What is the height of the final image of the tower if it is formed at 25 cm ? |
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Answer» Solution :Angular magnification of eyepiece = `(1 + (D)/(f_(e)) ) = 1 + (25)/(5 ) = 6 ` `therefore` Height of the final IMAGE = 6H = ` 6 XX 4.7 = 28.2 ` CM |
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