Question

Two converging lenses

(f₁ = 9.40 cm

and

f₂ = 5.50 cm)

are separated by 17.9 cm. The lens on the left has the longer focal length. An object stands 12.4 cm to the left of the left-hand lens in the combination. Locate the final image relative to the lens on the right. Obtain the overall magnification. Is the final image real or virtual? With respect to the original object, is the final image upright or inverted and is it larger or smaller?

(a) Locate the final image relative to the lens on the right.
cm  ---Select--- to the right of the right-hand lens to the left of the right-hand lens

(b) Obtain the overall magnification.


(c) Is the final image real or virtual?

realvirtual    

(d) Is the final image upright or inverted?

uprightinverted    

(e) Is it larger or smaller?

largersmaller   

Answer 1

For the first lens
1/f₁ = 1/d_o + 1/d_i
1/9.4 = 1/12.4 + (1/d_i)
d_i = 38.85 cm
Now the distance of the right lens from this image will be
= 38.85 - 17.9 = 20.95 cm
This image will work as object for the second lens
Hence for the second lens , object distance will be d₀₂ = -20.95 cm
focal length of second lens (f₂) = 5.5 cm
(1/f₂) = 1/d_o2 + 1/d_i2
1/5.5 = -(1/20.95) + (1/d_i2)
d_i2 = 4.356 cm
(a)Since the image distance is +ve therefore the image will be 4.356 cm toward right of the right lens.
(b) Magnification by the first lens
m₁ = -(d_i/d_o) = -(38.85/12.4) = -3.133
magnification by the second lens
m₂ = - (d_i2/d_o2) = -[4.356/(-20.95)] = 0.2079
Overall magnification = m₁*m₂ = -3.133*0.2079 = -0.6514
(c) Since the image distance for the second lens is +ve therefore the image will be real.
(d) Since the overall magnification is -ve therefore the image will be inverted.
(e) Since the overall magnification is less than 1 therefore the final image will be smaller.