+91-87964 74404
FlagBACK
Question icon
Grade 12th passIIT JEE Entrance Exam

A bi-convex lens is formed with two thin plano-convex lenses as shown in the figure. Refractive indexn 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 = 14 cm. For this bi-convex lens, for an object distance of 40 cm, the image distance will be

Profile image of lokesh
11 Years agoGrade 12th pass
Answers icon

1 Answer

Profile image of Vikas TU
6 Years ago
Given refractive index of lens 1 n₁ = 1.5 and  
refractive index of lens 1 n₂ = 1.2
Radius of curvature R₁ = 14 cm
Object distance u = - 40 cm
We know,
(1/f) = (μ -1) ( 1/R₁ - 1/R₂)
Thus the combined focal length will be
(1/f) = (1.5 -1)(1/14 - 1/∞) +(1.2 - 1)( 1/14 - 1/∞)
(1/f) = (0.5/14)+(0.2/14)
f =20 cm
From lens formula we have,
(1/v) - (1/u) =(1/f)
(1/v) = (1/f) + (1/u)
(1/v) = (1/20) -(1/40)
v = 40 cm  

Other Related Questions on iit jee entrance exam

question mark

Let S={x:x belongs to R(√3+√2)^(x^2-4) - (√3-√2)^(x^2-4)=10}. Then n(S) is equal to-

iit jee entrance exam
1 Answer Available

Last Activity: 10 Months ago

question mark
My rank is 29k, General, should I opt for any state college cse branch or some Nits in which in which I am getting ECE , chemical, mechanics etc....
 
iit jee entrance exam
1 Answer Available

Last Activity: 3 Years ago

question mark
How is Ranquer Eduzone,comparing, Akash,Fitjee,Resonance,Naryana
iit jee entrance exam
1 Answer Available

Last Activity: 4 Years ago

question mark
Can class 9th students study syllabus class  of 11th 
iit jee entrance exam
3 Answers Available

Last Activity: 4 Years ago

question mark
Sum of series 2[n/2] ! [n/2]!/ n! [2Co-2C2+3C2....+[-1]^n[n+1]2Cn] where n is an even positive integer. Kindly provide solution of this question and how to generalise this. How do we know that which expansion will give this?
iit jee entrance exam
1 Answer Available

Last Activity: 4 Years ago