Question

An equi – convex lens of glass, refractive index 1.5 has focal length 0.4 m in air. Calculate the

focal length of the lens when is immersed in a liquid of refractive index 1.7. What is the nature

of the lens in the liquid?​

Answer 1

Answer:

A thin lens of refractive index 1.5 has a focal length 15 cm in air. When the lens is placed in a medium of refractive index 4/3, what is the new focal length?

we can use the lens makers formula to solve this problem. Lens makers formula is 1/f=(n−1)(1/R1–1/R2) . Substituting the given values, the formula becomes

For the medium of refractive index 1.5

1/15=(1.5–1)(1/R1–1R2) ————(1)

For the medium of refractive index 4/3

1/f=(1.33–1)(1/R1–1/R2) ————(2)

Dividing (1) by (2) will yield the value of f 22.72 cm.