1. You align 2 lenses together with known focal lengths fi and fa. The equation to deter- mine the equivalent focal length fq is: D.???. ointaining the distance Solution Option (D) 1/f(eq) = 1/f1 + 1/f2, is the correct answer. Explanation - To prove the relation, we consider two thin convex lenses L1 and L2 in contact. Let the foci of the two lenses be f1 and f2 respectively. A point object O is placed. We assume a ray of light from O passing through the first lens L1. It forms an image at I1, in the absence of L2. Then we consider the image formation for the second lens. For the second lens the object is I1. I1 acts as a virtual object fro L2 and the image is formed at I. Let: u = object distance v = distance of image I v1 = distance of image I1 Using lens formula, we know that 1/f = 1/v + 1/u ------------------------------(i) Let us consider each lens separately: Applying sign convention For L1:- f = +f1 v = +v1 u = - u Now substituting them in equation (i), we get: 1/f1 = 1/ (+v1) + 1/ (-u) 1/f1 = 1/v1 -1/u -----------------------? (ii) For L2:- f = +f2 v = +v u = -v1 Now substituting the above in equation (i) we get: 1/f2 = 1/ (v) + 1/ (- v1) 1/f2 = 1/ (v) - 1/ (v1) ------------------------? (iii) Adding the equation (ii) and (iii): 1/f1 +1/f2 = [1/v1 -1/u] + [1/ (v) - 1/ (v1)] 1/f1 +1/f2 = 1/(v1) -1/(u) + 1/ (v) - 1/ (v1) 1/f1 +1/f2 = -1/u + 1/ (v) 1/f1 +1/f2 = 1/v -1/u ------------------------? (iv) Now, if the focal length of the combination of the two lenses L1 and L2 is F and the real image is formed at I (at v distance from the lens) and the object being at O (at distance u from the lens), then: 1/F = 1/v + 1/ (-u) 1/F = 1/v -1/u ----------------------? (v) From (iv) and (v), we get: 1/F = 1/f1 + 1/f2 means - 1 / f(eq) = 1/f1 + 1/f2. This is the relation we have mentioned above. .

1. 1. You align 2 lenses together with known focal lengths fi and fa. The equation to deter- mine the
equivalent focal length fq is: D.???. ointaining the distance
Solution
Option (D) 1/f(eq) = 1/f1 + 1/f2, is the correct answer.
Explanation -
To prove the relation, we consider two thin convex lenses L1 and L2 in contact. Let the foci of
the two lenses be f1 and f2 respectively. A point object O is placed. We assume a ray of light
from O passing through the first lens L1. It forms an image at I1, in the absence of L2. Then we
consider the image formation for the second lens. For the second lens the object is I1. I1 acts as a
virtual object fro L2 and the image is formed at I.
Let:
u = object distance
v = distance of image I
v1 = distance of image I1
Using lens formula, we know that
1/f = 1/v + 1/u ------------------------------(i)
Let us consider each lens separately:
Applying sign convention
For L1:-
f = +f1
v = +v1
u = - u
Now substituting them in equation (i), we get:
1/f1 = 1/ (+v1) + 1/ (-u)

2. 1/f1 = 1/v1 -1/u -----------------------? (ii)
For L2:-
f = +f2
v = +v
u = -v1
Now substituting the above in equation (i) we get:
1/f2 = 1/ (v) + 1/ (- v1)
1/f2 = 1/ (v) - 1/ (v1) ------------------------? (iii)
Adding the equation (ii) and (iii):
1/f1 +1/f2 = [1/v1 -1/u] + [1/ (v) - 1/ (v1)]
1/f1 +1/f2 = 1/(v1) -1/(u) + 1/ (v) - 1/ (v1)
1/f1 +1/f2 = -1/u + 1/ (v)
1/f1 +1/f2 = 1/v -1/u ------------------------? (iv)
Now, if the focal length of the combination of the two lenses L1 and L2 is F and the real image is
formed at I (at v distance from the lens) and the object being at O (at distance u from the lens),
then:
1/F = 1/v + 1/ (-u)
1/F = 1/v -1/u ----------------------? (v)
From (iv) and (v), we get:
1/F = 1/f1 + 1/f2
means -
1 / f(eq) = 1/f1 + 1/f2.
This is the relation we have mentioned above.