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A convex meniscus thin lens is made with the radius of curvature of the first convex surface being 25.0 cm and the second convex (towards object) surface 45.0 cm. If the glass used has index of refraction 1.500, what is the focal length of this lens?

Respuesta :

hopemichael95

Answer:

112.5cm

Explanation:

Using the formula

1/f = (n-1)(1/R1 + 1/R2)

1/f = (1.5 - 1)(1/25 - 1/45)

1/f = 0.5 x (4/225) = 2/225

f = 225/2

= 112.5cm

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