Question

An insect 3.00 mm tall is placed 20.0 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.

Calculate the location of the image this lens forms of the insect. in cm

Calculate the size of the image. in mm

Answer 1

Answer 2

Step 1: Calculate the focal length ($f$) of the lens using the lensmaker's formula:

The lensmaker's formula for a plano-convex lens (where one surface is flat) simplifies to:

$$\frac{1}{f}=(n-1)\left(\frac{1}{R_2}\right)$$

Substitute the given values:

$$\frac{1}{f}=(1.70-1)\left(\frac{1}{12.0}\right)=0.70\times \frac{1}{12.0}=\frac{0.70}{12.0}$$$$\frac{1}{f}=\frac{7}{120}\Rightarrow f=\frac{120}{7}\approx 17.14\text{ cm}$$

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Step 2: Calculate the image distance ($d_i$) using the thin lens formula:

$$\frac{1}{f}=\frac{1}{d_o}+\frac{1}{d_i}$$

Rearrange to solve for $d_i$:

$$\frac{1}{d_i}=\frac{1}{f}-\frac{1}{d_o}=\frac{7}{120}-\frac{1}{20}=\frac{7}{120}-\frac{6}{120}=\frac{1}{120}$$$$d_i=120\text{ cm}$$

The positive sign indicates the image is formed 120 cm to the right of the lens.

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Step 3: Calculate the size of the image ($h_i$) using the magnification formula:

Magnification ($m$) is given by:

$$m=-\frac{d_i}{d_o}=-\frac{120}{20}=-6$$

The negative sign indicates the image is inverted. The height of the image is:

$$h_i=m\times h_o=-6\times 3.00\text{ mm}=-18.0\text{ mm}$$

The magnitude of the image size is 18.0 mm.

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 Answers is :-

1. Location of the image: $\boxed{{\displaystyle 120\text{ cm}}}$ to the right of the lens.

2. Size of the image: $\boxed{{\displaystyle 18.0\text{ mm}}}$ (inverted)