Respuesta :
Answer:
Explanation:
For lens A
object distance u = - 13.1 cm , focal length f = 6.19 cm
From lens formula
1/v - 1/u = 1/f
1 / v + 1/13.1 = 1/6.19
1/v = 1/6.19 - 1/13.1
= .16155 - .07633
= .08522
v = 11.7 3 cm
For lens B
object distance u = - ( 55.7 - 11.73) = - 43.97 cm , focal length f = 27.9 cm
From lens formula
1/v - 1/u = 1/f
1 / v + 1/43.97 = 1/27.9
1/v = 1/27.9 - 1/43.97
= .03584 - .022742
= .013098
v = 76.35 cm
Image will be formed 76.35 cm behind lens B .
magnification of lens system
= m₁ x m₂ , m₁ is magnification by lens A and m₂ is magnification by lens B
= (11.73 / 13.1) x (76.35 / 43.97)
= .8954 x 1.73
= 1.5547
size of image = total magnification x size of object
= 1.5547 x 6.47
= 10 cm approx. The first image will be real and inverted and second image will be erect with respect to object.
The equation of the constructor of geometric optics allows to find the results for the distance to the image and its characteristics are:
a) The distance to the image is: q = 76.35 cm
b) The size of the image is: h ’= 1.007 cm
c) the image is real and straight with respect to the object
Given parameters
- Firefly height h = 6.47 mm
- Object distance to lens A p = 131 cm
- Lens focal length A fa = 6.19 cm
- Lens B focal length fb = 27.9 cm.
- Distance between lenses d = 55.7 cm.
To find
a) Distance from image to lens B
b) Image height
c) Image type formed by B
Geometric optics studies the position of the image formed by lenses and mirrors, for this it uses the equation of the constructor.
[tex]\frac{1}{f} = \frac{1}{p} + \frac{1}{q}[/tex]
Where f is the focal length, p and q are the distance to the object and the image respectively.
a) Let's apply the constructor'ss equation to each lens, remember that the distance is measured with respect to the lens.
Lens A
We look for the distance to the image.
[tex]\frac{1}{q_a} = \frac{1}{f_a} - \frac{1}{p} \\\frac{1}{q_a} = \frac{1}{6.19} - \frac{1}{13.1} \\\frac{1}{q_a} = 0.085215[/tex]
[tex]q_a[/tex] = 11.735 cm
Lens B
The distance to the object of lens B is the distance that separates the lenses minus the distance to the image, see attached.
[tex]p_b = d -q_a \\p_b = 55.7 -11.735[/tex]
[tex]p_b[/tex] = 43.965 cm
We use the constructor's equation to find the image formed by lens B
[tex]\frac{1}{q} = \frac{1}{f_b} - \frac{1}{p_b} \\\frac{1}{q} = \frac{1}{27.9} - \frac{1}{43.965} \\\frac{1}{q} = 0.013097[/tex]
q = 76.35 cm
b) the height of the image
The magnification of a lens is
m = [tex]\frac{h'}{h} = - \frac{q}{p}[/tex]
The magnification of several lenses is the product of the magnification of each lens.
[tex]m_{total} = m_a m_b \\m_{total} = \frac{11.735}{13.1} \ \frac{76.35}{43.965} \\m_{total} = 1.556[/tex]
We look for the height of the image
m = [tex]\frac{h' }{h}[/tex]
h ’= h m
h '= 0.647 1.556
h ‘= 1.007 cm
c) image type
There are some types of images:
- Real. If light rays pass through it.
- Virtual. If the prolongation of the rays form the image.
- Right. If the image has the same direction as the object.
- Inverted. If the image is inverted with respect to the direction of the object.
In this case, let's analyze what happens in each lens and then analyze the relationship between the initial object and the final image.
For lens A the image formed is real and inverted.
For lens B the object is real and the image is real and inverted.
For the relationship between the initial object and the final image.
The image is real because the rays pass through it.
The image on the right with respect to the object since they both have the same direction.
In conclusion, using the constructor's equation of the geometric optics we can find the results for the distance to the image and its characteristics are:
a) The distance to the image is: q = 76.35 cm
b) The size of the image is: h ’= 1.007 cm
c) the image is real and straight with respect to the object
Learn more here: brainly.com/question/13102239
