As We know that,
\(P=\frac{1}{f}=(µ-1)\bigg(\frac{1}{R_1}-\frac{1}{R_2}\bigg)\)
\(L_1:\frac{1}{f}=(µ-1)\bigg(\frac{1}{R_1}-\frac{1}{R_2}\bigg)=P_1=(µ-1)\bigg(\frac{2}{R}\bigg)=P\)
\(L_2:\frac{1}{f}=(µ-1)\bigg(\frac{1}{R_1}\bigg)=P_2=\frac{(µ-1)}{R}\)
\(L_3:\frac{1}{f}=(µ-1)\bigg(-\frac{1}{R_2}\bigg)=P_3=\frac{(µ-1)}{R}\)
Hence, Correct option is (A) : Power of \(L_1 = \frac{P}{2}\)
A bob of mass \(m\) is suspended at a point \(O\) by a light string of length \(l\) and left to perform vertical motion (circular) as shown in the figure. Initially, by applying horizontal velocity \(v_0\) at the point ‘A’, the string becomes slack when the bob reaches at the point ‘D’. The ratio of the kinetic energy of the bob at the points B and C is:
Let $ P_n = \alpha^n + \beta^n $, $ n \in \mathbb{N} $. If $ P_{10} = 123,\ P_9 = 76,\ P_8 = 47 $ and $ P_1 = 1 $, then the quadratic equation having roots $ \alpha $ and $ \frac{1}{\beta} $ is:
Lenses that are made by combining two spherical transparent surfaces are called spherical lenses. In general, there are two kinds of spherical lenses. Lenses that are made by joining two spherical surfaces that bulge outward are convex lenses, whereas lenses that are made by joining two spherical surfaces that curve inward are concave lenses.