16 cm
12 cm
To find the focal length of the lens, we use the lens formula:
\(\frac{1}{f} = \frac{1}{v} - \frac{1}{u}\)
where \(f\) is the focal length, \(v\) is the image distance, and \(u\) is the object distance. Given that the object distance \(u = -20\) cm (convention: object distance is negative) and the screen is 50 cm away from the object, the image distance \(v = 20 + 50 = 70\) cm.
Substitute these values into the lens formula:
\( \frac{1}{f} = \frac{1}{70} - \frac{1}{-20} \)
\( \frac{1}{f} = \frac{1}{70} + \frac{1}{20} \)
Convert to a common denominator:
\( \frac{1}{f} = \frac{20 + 70}{1400} = \frac{90}{1400} \)
Therefore, \( f = \frac{1400}{90} = \frac{140}{9} \approx 15.56 \text{ cm} \)
So, the focal length of the lens is 16 cm.
| S. No. | Particulars | Amount (in ₹ crore) |
|---|---|---|
| (i) | Operating Surplus | 3,740 |
| (ii) | Increase in unsold stock | 600 |
| (iii) | Sales | 10,625 |
| (iv) | Purchase of raw materials | 2,625 |
| (v) | Consumption of fixed capital | 500 |
| (vi) | Subsidies | 400 |
| (vii) | Indirect taxes | 1,200 |
Draw a rough sketch for the curve $y = 2 + |x + 1|$. Using integration, find the area of the region bounded by the curve $y = 2 + |x + 1|$, $x = -4$, $x = 3$, and $y = 0$.