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

(A) therefore the image is

63 cm to the right of the lensthe image size is -13.22 cmit is realit is inverted

(B) therefore the image is

63 cm to the right of the lensthe image size is -13.22 cmit is realit 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

where

s' = distance of the imagef = focal lengthbut we first need to find the focal length before we can apply this formula

f =

f = 17.9 cm

now that we have the focal length we can apply

=\frac{1}{s'}

=\frac{1}{s'}

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

magnification =   where y' is the height of the image, therefore

y' = x 0.525 = -13.22 cm

therefore the image is

63 cm to the right of the lensthe image size is -13.22 cmit is realit 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

where

s' = distance of the imagef = focal lengthbut we first need to find the focal length before we can apply this formula

f =

f = 17.9 cm

now that we have the focal length we can apply

=\frac{1}{s'}

=\frac{1}{s'}

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

magnification =   where y' is the height of the image, therefore

y' = x 0.525 = -13.22 cm

therefore the image is

63 cm to the right of the lensthe image size is -13.22 cmit is realit is inverted

1) Yes

2)

Explanation:

1)

To solve this part, we have to calculate the pressure at the depth of the batyscaphe, and compare it with the maximum pressure that it can withstand.

The pressure exerted by a column of fluid of height h is:

where

is the atmospheric pressure

is the fluid density

is the acceleration due to gravity

h is the height of the column of fluid

Here we have:

is the sea water density

h = 5440 m is the depth at which the bathyscaphe is located

Therefore, the pressure on it is

Since the maximum pressure it can withstand is 60 MPa, then yes, the bathyscaphe can withstand it.

2)

Here we want to find the force exerted on the bathyscaphe.

The relationship between force and pressure on a surface is:

where

p is hte pressure

F is the force

A is the area of the surface

Here we have:

is the pressure exerted

The bathyscaphe has a spherical surface of radius

r = 3 m

So its surface is:

Therefore, we can find the force exerted on it by re-arranging the previous equation: