Q55E - University Physics with Modern Physics

Question

If the person in Exercise 34.55 chooses ordinary glasses over contact glasses, what power lens (in dioptres) does she need to correct her vision if the lenses are 2.0 cm in front of the eye?

Step-by-Step Solution

Answer

She needs -1.37 dioptres power to correct her vision.

Step 1: Object-Image Relationship and sign rules for the variables

                   $\frac{\mathbf{1}}{\mathbf{s}}\mathbf{+}\frac{\mathbf{1}}{\mathbf{s}}\mathbf{=}\frac{\mathbf{1}}{\mathbf{f}}$

Where s = object distance from the lens

           s' =  image distance from the lens

            f = focal length of the lens

Sign rules for the variables:

1. Sign rule for the object distance: when the object is on the same side of the refracting surface as the incoming light, object distance is positive; otherwise, it is negative.
2. Sign rule for the image distance: when the image is on the same side of the refracting surface as the outgoing light, image distance is positive; otherwise, it is negative.

Step 2: Calculate the power required to correct her vision

To make the person see object beyond 75 cm (objects at infinity), we need to put glasses to make an image of objects at infinity at 75 cm from the eye

                (75cm - 2cm  = 73cm) -> from the glasses

Here, => the far point which the eyes see without the lens

        $\mathrm{s}=\infty$ => the far point which the eyes see with the lens

Use the formula, 

                                $\frac{1}{\mathrm{s}}+\frac{1}{\mathrm{s}\text{'}}=\frac{1}{\mathrm{f}}$

                               $\frac{1}{\infty }+\frac{1}{-73}=\frac{1}{\mathrm{f}}$

                             $\Rightarrow \mathrm{f}=-73\mathrm{cm}=-0.73\mathrm{m}$

Therefore, power = 1/focal length $=\frac{1}{-0.73\mathrm{m}}=-1.37$dioptres

Thus, she needs -1.37 dioptres power to correct her vision.