Question

A point source of light illuminates an aperture 1.70 m away. A 11.0 cm -wide bright patch of light appears on a screen 0.800 m behind the aperture. Part A How wide is the aperture

Answer 1

Answer:

7.48 cm

Explanation:

The diagrammatic representation of information is shown in the image below:

From the image, Δ X₁X₂ and Δ Y₁Y₂ are equivalent triangles.

where;

X₁X₂ = width of the aperture

Using equivalent triangles, It implies that:

$\implies \dfrac{X_1X_2}{Y_1Y_2} = \dfrac{1.70}{1.70+0.800} \\ \\ \implies X_1X_2=\dfrac{1.70}{1.70+0.800}\times 11.0$

$X_1X_2= \dfrac{1.70}{2.50}\times 11.0 cm$

X₁X₂ = 7.48 cm