Physics, asked by manojkumarbannu23, 9 months ago

32. An equiconvex lens is made of glass of
refractive index 1.5 and has a focal length
of 10 cm in air. The lens is cut into two equal
halves along a plane perpendicular to its
principal axis to yield two plano-convex
lenses. The two pieces are glued such that
the convex surfaces touch each other. If
this combiantion lens is immersed in water
of refractive index 4/3 its focal length (in
cm) is​

Answers

Answered by Anonymous
1

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Answered by sonuvuce
2

If the combination of lens is immersed in water, its focal length is 40 cm

Explanation:

Given

The focal length of the lens f' = 10 cm[tex]</p><p>Since this lens is cut into two equal part, the focal length of each part will be</p><p>[tex]=2f'=1\times10=20 cm

We know that when two lenses of focal length f_1 and f_2 are kept in contact, then the effective focal length is given by

\boxed{\frac{1}{f}=\frac{1}{f_1}+\frac{1}{f_2}}

\implies \frac{1}{f}=\frac{1}{20}+\frac{1}{20}

\implies\frac{1}{f}=\frac{1}{10}

\implies f=10 cm

Therefore, when the two pieces are glued together, there is no change in focal length

Given the refractive index of glass w.r.t. air _g\mu_a=1.5

Refractive index of water w.r.t. air _w\mu_a=\frac{4}{3}

Refractive index of glass w.r.t. water _g\mu_w

_g\mu_w=\frac{_g\mu_a}{_w\mu_a}

\implies _g\mu_w=\frac{1.5}{4/3}

\implies _g\mu_w=\frac{4.5}{4}

if focal length of the lens in water is f_w then

\boxed{f_w=(\frac{_g\mu_a-1}{_g\mu_w-1})f}

\implies f_w=(\frac{1.5-1}{(4.5/4)-1})\times 10

\implies f_w=\frac{0.5\times 4}{0.5}\times 10

\implies f_w=40 cm

Hope this answer is helpful.

Know More:

Q: A glass convex lens of refractive index 1.5 has focal length 8 cm when in air what would be focal length of lens when immersed in water of refractive index 1.33.

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