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 \(L_1 = \frac{P}{2}\)
Let one focus of the hyperbola $ \frac{x^2}{a^2} - \frac{y^2}{b^2} = 1 $ be at $ (\sqrt{10}, 0) $, and the corresponding directrix be $ x = \frac{\sqrt{10}}{2} $. If $ e $ and $ l $ are the eccentricity and the latus rectum respectively, then $ 9(e^2 + l) $ is equal to:
The largest $ n \in \mathbb{N} $ such that $ 3^n $ divides 50! 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.