A bi-convex lens is formed with two thin plano-convex lenses as shown in the figure. The refractive index n of the first lens is 1.5 and that of the second lens is 1.2 . Both the curved surfaces are of the same radius of curvature, R=14cm . For this bi-convex lens, for an object distance of 40cm , the image distance will be,

Question

A bi-convex lens is formed with two thin plano-convex lenses as shown in the figure. The refractive index n of the first lens is 1.5 and that of the second lens is 1.2 . Both the curved surfaces are of the same radius of curvature, R=14cm . For this bi-convex lens, for an object distance of 40cm , the image distance will be, <img class="img-fluid question-image" alt="Question" src="https://cdn.tardigrade.in/q/nta/p-pi1hmjpb0jdzb0ri.jpg" /> (A) -280.0cm (B) 40.0cm (C) 21.5cm (D) 13.3cm

Tardigrade Admin · Accepted Answer

(1/f1)=(1 . 2 - 1)((1/∞ ) - (1/- 14))=(0 . 2/14) (1/f2)=(1 . 5 - 1)((1/14) - (1/∞ ))=(0 . 5/14) (1/feq)=(0 . 7/14)=(1/20) (1/v)-(1/- 40)=(1/20) v=40cm .

$- 280.0 c m$

$40.0 c m$

$21.5 c m$

$13.3 c m$

$\frac{1}{f _{1}} = (1.2 - 1) (\frac{1}{\infty} - \frac{1}{- 14}) = \frac{0.2}{14}$
$\frac{1}{f _{2}} = (1.5 - 1) (\frac{1}{14} - \frac{1}{\infty}) = \frac{0.5}{14}$
$\frac{1}{f _{e q}} = \frac{0.7}{14} = \frac{1}{20}$
$\frac{1}{v} - \frac{1}{- 40} = \frac{1}{20}$
$v = 40 c m$ .

−280.0cm

40.0cm

21.5cm

13.3cm

f1​1​=(1.2−1)(∞1​−−141​)=140.2​ f2​1​=(1.5−1)(141​−∞1​)=140.5​ feq​1​=140.7​=201​ v1​−−401​=201​ v=40cm .

$- 280.0 c m$

$40.0 c m$

$21.5 c m$

$13.3 c m$

$\frac{1}{f _{1}} = (1.2 - 1) (\frac{1}{\infty} - \frac{1}{- 14}) = \frac{0.2}{14}$
$\frac{1}{f _{2}} = (1.5 - 1) (\frac{1}{14} - \frac{1}{\infty}) = \frac{0.5}{14}$
$\frac{1}{f _{e q}} = \frac{0.7}{14} = \frac{1}{20}$
$\frac{1}{v} - \frac{1}{- 40} = \frac{1}{20}$
$v = 40 c m$ .