Question

An insect 5.25 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 12.5 cm , and the index of refraction of the lens material is 1.70. a)Calculate the location and size of the image this lens forms of the insect. Is it real or virtual? Erect or inverted?
b)Repeat part (a) if the lens is reversed

Answer 1

Answer:

(A) therefore the image is

• 63 cm to the right of the lens
• the image size is -13.22 cm
• it is real
• it is inverted

(B) therefore the image is

• 63 cm to the right of the lens
• the image size is -13.22 cm
• it is real
• it is inverted

Explanation:

height of the insect (h) = 5.25 mm = 0.525 cm

distance of the insect (s) = 25 cm

radius of curvature of the flat left surface (R1) = ∞

radius of curvature of the right surface (R2) = -12.5 cm (because it is a planoconvex lens with the radius in the direction of the incident rays)

index of refraction (n) = 1.7

(A) we can find the location of the image by applying the formula below

$\frac{1}{f} =\frac{1}{s'} +\frac{1}{s}$ where

• s' = distance of the image
• f = focal length

• but we first need to find the focal length before we can apply this formula

$\frac{1}{f} =(n-1)(\frac{1}{R1} -\frac{1}{R2} )$

$\frac{1}{f} =(1.7-1)(\frac{1}{∞} -\frac{1}{-12.5} )$

$\frac{1}{f} =(0.7)(0 + \frac{1}{12.5} )$

$\frac{1}{f} =\frac{0.7}{12.5}$

f = $\frac{12.5}{0.7}$

f = 17.9 cm

now that we have the focal length we can apply $\frac{1}{f} =\frac{1}{s'} +\frac{1}{s}$

$\frac{1}{f} - \frac{1}{s}$ =\frac{1}{s'}

$\frac{1}{17.9} - \frac{1}{25}$ =\frac{1}{s'}

$\frac{25 - 17.9}{17.9 x 25} =\frac{1}{s'}$

$\frac{7.1}{447.5} =\frac{1}{s'}$

s' = \frac{447.5}{7.1}[/tex]  = 63 cm to the right of the lens

magnification =$\frac{-s'}{s} =\frac{y'}{y}$   where y' is the height of the image, therefore

$\frac{-s'}{s} =\frac{y'}{y}$

$\frac{-63}{25} =\frac{y'}{52.5}$

y' = $\frac{-63}{25}$ x 0.525 = -13.22 cm

therefore the image is

• 63 cm to the right of the lens
• the image size is -13.22 cm
• it is real
• it is inverted

(B) if the lens is reversed, the radius of curvatures would be interchanged

radius of curvature of the flat left surface (R1) = ∞

radius of curvature of the right surface (R2) = 12.5 cm

we can find the location of the image by applying the formula below

$\frac{1}{f} =\frac{1}{s'} +\frac{1}{s}$ where

• s' = distance of the image
• f = focal length

• but we first need to find the focal length before we can apply this formula

$\frac{1}{f} =(n-1)(\frac{1}{R1} -\frac{1}{R2} )$

$\frac{1}{f} =(1.7-1)(\frac{1}{12.5} -\frac{1}{∞} )$

$\frac{1}{f} =(0.7)( \frac{1}{12.5} - 0)$

$\frac{1}{f} =\frac{0.7}{12.5}$

f = $\frac{12.5}{0.7}$

f = 17.9 cm

now that we have the focal length we can apply $\frac{1}{f} =\frac{1}{s'} +\frac{1}{s}$

$\frac{1}{f} - \frac{1}{s}$ =\frac{1}{s'}

$\frac{1}{17.9} - \frac{1}{25}$ =\frac{1}{s'}

$\frac{25 - 17.9}{17.9 x 25} =\frac{1}{s'}$

$\frac{7.1}{447.5} =\frac{1}{s'}$

s' = \frac{447.5}{7.1}[/tex]  = 63 cm to the right of the lens

magnification =$\frac{-s'}{s} =\frac{y'}{y}$   where y' is the height of the image, therefore

$\frac{-s'}{s} =\frac{y'}{y}$

$\frac{-63}{25} =\frac{y'}{52.5}$

y' = $\frac{-63}{25}$ x 0.525 = -13.22 cm

therefore the image is

• 63 cm to the right of the lens
• the image size is -13.22 cm
• it is real
• it is inverted

Answer 2

The location and size of image of insect is 63 cm right to the lens and -13.22 mm respectively. The image is real and inverted.  

What is focal length of the lens?

The focal length of the lens is the length of the distance between the middle of the lens to the focal point.

It can be found out using the following formula as,

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

Here, (v)is the distance of the image, (u) is the distance of the object, and (f) is the focal length of the lens.

• a) The location and size of the image the lens forms of the insect-

The focal length of the lens can also be given as,

$\dfrac{1}{f}=(n-1)\left(\dfrac{1}{R_1}-\dfrac{1}{R_2}\right)$

Here, n is the index of lens material and R(1), R(2) is the radius of the curvature.

The radius of curvature is 12.5 cm and index of refraction of the lens material is 1.70. Put the values,

$\dfrac{1}{f}=(1.7-1)\left(\dfrac{1}{\infty}-\dfrac{1}{12.5}\right)\\f=17.9\rm cm$

The insect is placed 25.0 cm to the left of a thin planoconvex lens and the focal length of this lens is 17.9 meters. Thus the distance of the image by the lens formulas is,

$\dfrac{1}{v}+\dfrac{1}{25}=\dfrac{1}{17.9}\\v=63\rm cm$

Let the height of the image is x. By the magnification formula,

$\dfrac{-63}{25}=\dfrac{x}{5.25}\\x=-13.22 \rm mm$

Thus, the location and size of image of insect is 63 cm right to the lens and -13.22 mm respectively. The image is real and inverted.  

b) The location and size of the image the lens forms of the insect, when the lens reversed-

By reversing the lens, the radius of curvature of both the surface is interchanged, but the result remain same.

Thus, by reversing the lens, the location and size of image of insect is 63 cm right to the lens and -13.22 mm respectively. The image is real and inverted.  

Learn more about the focal length here;

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