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The near and far points of a person are at 40 cm and 250 cm respectively. Find the power of the lens he/she should use while reading at 25cm. With this lens on the eye, what maximum distance is clearly visible? |
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Answer» Solution :If an object is placed at 25 cm from the correcting lens, it should produce the virtual image at 40 cm. Thus, `u=-25CM,v=-40,1/f=1/v-1/u=-1/(40cm)+1/(25cm)` or, `f=200/3cm=+2/3morP=1/f=+1.5D` The UNAIDED eye can see a maximum DISTANCE of 250 cm. Suppose the maximum distance for clear vision is d when the lens is used. Then the object at a distance d is imaged by the lens at 250 cm. We have, `1/v-1/u=1/for-1/(250CM)-1/d=3/(200cm)ord=-53cm` Thus, the person will be ABLE to see upto a maximum distance of 53 cm. |
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