For distant vision = −0.181 m, for near vision = 0.667 m
The power P of a lens of focal length f is given by the relation.
\(\text{Power (P)} = \frac{1}{f}\)
(i) Power of the lens used for correcting distant vision = −5.5 D
Focal length of the required lens, \(f =\frac{1}{\text{P}}\)
\(f = \frac{1}{-5.5}\)
f = -0.181 m
The focal length of the lens for correcting distant vision is −0.181 m.
(ii) Power of the lens used for correcting near vision = +1.5 D
Focal length of the required lens, \(f =\frac{1}{\text{P}}\)
\(f = \frac{1}{1.5}\) = +0.667 m
The focal length of the lens for correcting near vision is 0.667 m.

In the adjoining figure, $\triangle CAB$ is a right triangle, right angled at A and $AD \perp BC$. Prove that $\triangle ADB \sim \triangle CDA$. Further, if $BC = 10$ cm and $CD = 2$ cm, find the length of AD. 
सड़क सुरक्षा के प्रति जागरूकता हेतु ट्रैफिक पुलिस की ओर से जनहित में जारी एक आकर्षक विज्ञापन लगभग 100 शब्दों में तैयार कीजिए।
The following data shows the number of family members living in different bungalows of a locality:
| Number of Members | 0−2 | 2−4 | 4−6 | 6−8 | 8−10 | Total |
|---|---|---|---|---|---|---|
| Number of Bungalows | 10 | p | 60 | q | 5 | 120 |
If the median number of members is found to be 5, find the values of p and q.