Question

A thin lens made of glass of refractive index 5 has a front surface +11D power and back surface -6D. If thin lens is submerged in a liquid of refractive index 1.6 , the resulting power of the lens is

please answer with explanation​

Answer 1

Answer:

The lens maker formula reads as follows .

1/f=[(n2-n1)/n1][(1/R1)-(1/R2)]. Here,

f is focal length of thin lens.

n2 is absolute refractive index of material of lens.

n1 is refractive index of medium surrounding the lens.

R1 is radius of curvature of the first surface of lens.(on which the ray is incident ).

R2 is radius of curvature of the second surface from where ray leaves the lens.

In Cartesian sign convention ,R1=15 cm, R2=-15 cm.

Also,n1=1.65 and n2=1.6 (given).

Substituting all these values in the above mentioned lens maker formula, we have,

(1/f)=(1.6–1.65)/(1.65)[(1/15)+(1/15)]=-[(0.05)/(1.65)](2/15)=-0.1/(1.65X15)

Therefore, f=-1.65X150=-247.5~-250cm

Explanation:

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