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Diameter of a Plano-convex lens is 6 cm and the thickness at the center is 3 mm. If the speed of light in the material of the lens is 2 x 10^8 m/sec, what is the focal length of the lens?
diameter = 2 Radius (R) 2R = 6 R = 3cm 1/f = (u-1)/R ......1 this is the relation bw focal length of plano convex lens & u(refractive index) ... u = c/v = 3*108/2*108 = 1.5 now put u in eq 1 , will get focal length
diameter = 2 Radius (R)
2R = 6
R = 3cm
1/f = (u-1)/R ......1
this is the relation bw focal length of plano convex lens & u(refractive index) ...
u = c/v = 3*108/2*108 = 1.5
now put u in eq 1 , will get focal length
radius will be 15 not 3! use geometry. let (x) be radius of curvature .BASE- (x-3) HEIGHT -(3) HYPOTANEOUS -(x) solve and get (x)! :)
from right angled ∆OAB,OA² = AB² + BO²⇒R² = (radius)² + (radius - thickness)²⇒R² = 3² + (R - 0.3)²⇒R² = 9 + R² - 0.6R + 0.09⇒0.6R = 9.009⇒R = 90.09/6 ≈ 15 cmit is given that, speed of light in material of lens is 2 × 10^ 8m/sso, refractive index, μ = 3 × 10^8/2 × 10^8 = 1.5now using formula, 1/f = (μ - 1)[1/R1 - 1/R2]here, R1 = 15cm , R2 = ∞so, 1/f = (1.5 - 1)[1/15 - 1/0]= 0.5/15= 1/30hence, f = 30 cm
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