Q.i) A person having long-sight cannot see things distinctly at a distance less than 40 cm. If he wants to see things situated at 25 cm from him, what should be the power of his spectacles? ii) Why is the objective of an Astronomical telescope made of large diameter? iii) Establish the relation between apparent depth and real depth with refractive index of dense medium.
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Start your 14-day free trial to unlock the full solution →(i) Power of the correcting convex lens = +1.5 D. (ii) A wide objective collects more light and resolves finer detail. (iii) Refraction of near-normal rays at a plane surface gives n = real depth / apparent depth, hence apparent depth = real depth / n.
(i) Hypermetropia (long-sight) correction. The person's near point is 40 cm; he wants to read an object placed at 25 cm. The spectacle lens must take the object at 25 cm and form a virtual, erect image at his near point 40 cm (so he can see it). Using the lens formula with sign convention (both distances on the same side, so negative):
u = -25 cm, v = -40 cm.
1/f = 1/v - 1/u = 1/(-40) - 1/(-25) = -1/40 + 1/25 = (-25 + 40)/1000 = 15/1000 = 3/200 per cm.
So f = 200/3 cm = 66.7 cm = 2/3 m (positive, a convex lens).
Power P = 1/f(in metres) = 1/(2/3) = +1.5 D.
(ii) Large-diameter objective of an astronomical telescope. A larger objective aperture (a) collects more light, so the image of a faint or distant object is brighter, and (b) increases the resolving power (which is proportional to the aperture diameter, since the limit of resolution about 1.22*lambda/D decreases as D increases), so finer angular detail can be distinguished.
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