Question

A lens of focal length f₁ = +20 cm has an object positioned 50 cm in front of it. A second lens of focal length f₂= -15 cm is located 10 cm behind the first lens. a) Determine the two image positions for this combination of lenses. b) Is each image real or virtual, upright or inverted, and magnified or demagnified?

Answer 1

First we will find the distance of image from first lens of f₁ = + 20cm,

we will denote distance of object in front of first lens as "s₁" and distance of its image by first lens as "s'₁" cm behind the first lens (on other side)

1/s₁ + 1/s'₁ = 1/f₁

s'₁ = (s₁ * f₁) /(s₁ - f₁) = (50*20)/(50-20) = 33.33 cm

so, Image of object by first lens is 33.33cm behind the first lens

Magnification, m₁ = -s'₁ /s₁ = - 33.33/50 = -0.67 (Negative value shows inverted image)

As the Second lens is 10 cm behing first lens; Image of first lens will be object of second lens. Image of first lens is 33.33 - 10 = 23.33cm behind the Second lens (Virtual Object) and the image is inverted as well

s₂ = -23.33cm

f₂ = -15cm

so, distance of image of second lens s'₂ will given as:

s'₂ = (s₂ *f₂) / (s₂ - f₂)

= (-23.33) * (-15) / (-23.33 -(-15))

s'₂ = -42.01cm (Negative value denotes that Image of object by second lens will be 42.01 cm on the front of Second Lens and the image is Virtual)

m₂ = - s'₂ / s₂ = - (-42.01)/(-23.33) = - 1.8

Total Magnification , m_T = m₁ *m₂ = -0.67 * (-1.8) = 1.2 (FInal Image is magnified by 1.2 times of object)

a) image of object by first lens is 33.33cm behind First lens (Real and Inverted Image)

Image of object by Both the lens combined is 42.01 cm in front of second lens (or 42.01 -10 = 32.01cm in front of first Lens)

b) Image by first Lens only is Real and Inverted.(magnication by first lens is 0.67 times)

Final image by both lenses are Virtual and Upright. It is magnified by 1.2 times of original object.