Question

A converging lens (f = 18.0 cm) is located 20.0 cm to the left a diverging lens (f = -25.0 cm). An object is placed 50.0 cm to the left of the converging lens.

Calculate the position of the final image relative to the diverging lens.

Give your answer to 1 decimal place.

Note: if the 1st decimal place is a 0, canvas might auto-remove this but it will still be counted.

Answer 1

The position of the final image relative to the diverging lens to 1 decimal place to 35.4 cm to the left side of the diverging lens.

A converging lens is a lens that puts together rays of light that pass through its primary axis.

A diverging lens is a lens that causes light rays to spread out after they pass through the optical axis of the lens.

By using the Lens Formula from the information given:

$\dfrac{1}{f}= \dfrac{1}{v}+\dfrac{1}{u}$

$-\dfrac{1}{25}= \dfrac{1}{v}+\dfrac{1}{50}$

$-\dfrac{1}{v}= \dfrac{1}{25}+\dfrac{1}{50}$

$-\dfrac{1}{v}= \dfrac{3}{50}$

v = 16.67 cm

However, the above image distance relates to the object of the diverging lens.

The object distance can now be expressed as;

u' = (20 + 16.67) cm

u' = 36.67 cm

Again for a diverging lens according to the lens formula;

The final image distance formed is:

$\dfrac{1}{18} = \dfrac{1}{v'}+\dfrac{1}{36.67}$

$\dfrac{1}{v'}= \dfrac{1}{18} -\dfrac{1}{36.67}$

$\dfrac{1}{v'}= \dfrac{1867}{6600}$

$v' = \dfrac{66006}{1867}$

v' = 35.4 cm

Therefore, we can conclude that the object is placed at a distance of 35.4 cm to the left of the diverging lens

Learn more about the lens here:

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