Question

An insect 5.00 mm tall is placed 25.0 cm to the left of a thin planoconvex lens. The left surface of this lens is flat, the right surface has a radius of curvature of magnitude 12.0 cm , and the index of refraction of the lens material is 1.70

Where is the image located in relation to the lens?
How much to the left of the lens is the object?
How far from the lens is the image?

Answer 1

Manufacturer equation.

$\frac{1}{f}=\left ( n-1 \right )\left ( \frac{1}{R_{1}}\dotplus \frac{1}{R_{2}} \right )$

$R_{1}=12cm$ side convex

$R_{2}=\infty$ side plano

$n=1.7$

We evaluated

$\frac{1}{f}=\left ( 1.7-1 \right )\left ( \frac{1}{12cm}\dotplus \frac{1}{\infty} \right )$

The focal distance is calculated

$\frac{1}{f}=\left ( 0.7 \right )\left ( \frac{1}{12cm} \right )\rightarrow f=\frac{12cm}{0.7}=17,1cm$

We use the thin lens equation.

$\frac{1}{di}\dotplus \frac{1}{do}=\frac{1}{f}$

Position the object.
Image position.
focal distance

As the object distance $d_{o}=25cm$

To the left of the slow, we find the expression for the distance of the image.

$\frac{1}{di}\dotplus \frac{1}{do}=\frac{1}{f}\Rightarrow \frac{1}{di}=\frac{1}{f}-\frac{1}{do}$

$\frac{1}{di}=\frac{do-f}{f\cdot do}$

investing.

$di=\frac{f\cdot do}{do-f}=\frac{17.1cm\cdot 25cm}{25cm-17.1cm}=54.1cm$   

$di=54.1cm$

Where is the image located in relation to the lens?

if the distance is positive image it is to the right of the lens.

How much to the left of the lens is the object?

The same problem gives you the answer, is 25 centimeters to the left of the lens.

How far from the lens is the image?

As we calculate the image is to 54.1cm the lens.