A person needs a lens of power –5.5 dioptres for correcting his distant vision. For correcting his near vision he needs a lens of power +1.5 dioptre. What is the focal length of the lens required for correcting (i) distant vision, and (ii) near vision?
Power of the lens is given (P) for distant vision = -5.5 D
Power of the lens (P2) is given for near vision = + 1.5 D
1. Focal length of the lens required for correcting lens of distant vision
Formula used, P = 1/ f where, P is the power of the lens
And, f is the focal length
To correct f
f = 1/ P
f = 1/ -5.5
f = - 0.1818 m.
2. Focal length of the lens required for correcting lens of near vision
Formula used, P = 1/ f where, P is the power of the lens
And, f is the focal length
To correct f
f = 1/ P
f = 1/ 1.5
f = - 0.667 m.
Samples of four metals A, B, C and D were taken and added to the following solution one by one. The results obtained have been tabulated as follows.
| Metal | Iron(II) sulphate | Cooper(II) sulphate | Zinc sulphate | Silver nitrate |
| A | No reaction | Displacement | ||
| B | Displacement | No reaction | Displacement | |
| C | No reaction | No reaction | No reaction | No reaction |
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Welcome to the NCERT Solutions for Class 10 Science - Chapter . This page offers a step-by-step solution to the specific question from Excercise 2 , Question 5: A person needs a lens of power –5.5 dioptres for correcting his distant vision. For correcting....
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