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Dvinal [7]
3 years ago
7

The radii of curvature of a biconvex lens are 4 cm and 15 cm. The lens is in air, its index of refraction is 1.5. An object is a

t 1 m before the front surface of the lens. Calculate the distance of the image from the back surface of the lens.
Physics
1 answer:
rewona [7]3 years ago
8 0

Answer:

v = 6.315 cm

Explanation:

given,

R₁ = 4 cm = 0.04 m

R₂ = 15 cm = 0.15 m

n =1.5

\dfrac{1}{f}=\dfrac{1}{v}-\dfrac{1}{u}=(n-1)(\dfrac{1}{R_1}-\dfrac{1}{R_2})

\dfrac{1}{v}-\dfrac{1}{-1}=(1.5-1)(\dfrac{1}{0.04}+\dfrac{1}{0.15})

\dfrac{1}{v}+1 = 0.5 \times 31.66

\dfrac{1}{v} = 15.833

v = 0.06315 m

v = 6.315 cm  

hence, the distance of the image from the back surface is v = 6.315 cm

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Explanation:

By First Law of Thermodynamics, we see that amount of thermal energy (Q), in joules, is equal to the change in internal energy. From statement we understand that change in internal energy consisting in two latent components (U_{l,ice}, U_{l,steam}), in joules, and two sensible component (U_{s,w}), in joules, that is:

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h_{f,w} - Latent heat of fussion of water, in joules per kilogram.

h_{v,w} - Latent heat of vaporization of water, in joules per kilogram.

c_{ice} - Specific heat of ice, in joules per kilogram per degree Celsius.

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