A thin equiconvex lens (μ=3 / 2) of focal length 10 cm is cut and separated and a material of refractive index 3 is filled between them. What is the focal length of the combination?

Question

A thin equiconvex lens (μ=3 / 2) of focal length 10 cm is cut and separated and a material of refractive index 3 is filled between them. What is the focal length of the combination? (A) -10 cm (B) -10 / 4 cm (C) -10 / 3 cm (D) None of these

Tardigrade Admin · Accepted Answer

<img class="img-fluid question-image" alt="image" src="https://cdn.tardigrade.in/img/question/physics/24a1f08ce831ea7112256ea42ef0decc-.png" /> (1/f)=(μ-1)((1/R1)-(1/R2)) R=10 cm f prime=(3-1)((1/-10)-(1/10)) f prime=(-10/4) (1/f eq )=(1/20)-(4/10)+(1/20)=(2/20)-(4/10) f eq =-10 / 3 cm

Q. A thin equiconvex lens $(μ = 3/2)$ of focal length $10 c m$ is cut and separated and a material of refractive index 3 is filled between them. What is the focal length of the combination?

$- 10 c m$

$- 10/4 c m$

$- 10/3 c m$

None of these

$\frac{1}{f} = (μ - 1) (\frac{1}{R _{1}} - \frac{1}{R _{2}})$
$R = 10 c m$
$f^{'} = (3 - 1) (\frac{1}{- 10} - \frac{1}{10})$
$f^{'} = \frac{- 10}{4}$
$\frac{1}{f _{e q}} = \frac{1}{20} - \frac{4}{10} + \frac{1}{20} = \frac{2}{20} - \frac{4}{10}$
$f_{e q} = - 10/3 c m$

Q. A thin equiconvex lens (μ=3/2) of focal length 10cm is cut and separated and a material of refractive index 3 is filled between them. What is the focal length of the combination?

−10cm

−10/4cm

−10/3cm