Question

) A convex-concave thin lens is made with the radius of curvature of the convex surface being 25.0 cm and the concave surface 45.0 cm. If the glass used has index of refraction 1.500, what is the focal length of this lens

Answer 1

Answer:

$f= +113\ cm$

Explanation:

Using Lens Maker's formula

$\frac{1}{f}=(n-1)\left(\frac{1}{R_{1}}-\frac{1}{R_{2}}\right)$

$R_{1}=+25 cm,\ R_{2}=+45 cm, n=1.5$

$\frac{1}{f}=(1.5-1)\left(\frac{1}{25}-\frac{1}{45}\right)$

$\frac{1}{f}=0.5 \times \frac{45-25}{45 \times 25}$

$\frac{1}{f}=0.5 \times \frac{20}{45 \times 25}$

$f = 112.5\ cm$

$f= +113\ cm$

Hence, focal length is equal to 113 cm.