Answer:
f = 55mm, h ’= -9.89 cm
f = 200 mm, h ’= 42.5 cm
Explanation:
For this exercise let's start by finding the distance to the image, using the equation of the constructor

where f is the focal length, p and q are the distances to the object and image, respectively
lens with f₁ = 55mm = 0.55cm
=
= 1.718
q₁ = 0.582 m
lens with f₂ = 200mm = 2m
=
= 0.4
q₂ = 2.5 m
the magnification of a lens is given by
m =
h ’=
let's calculate for each lens
f = 55mm
h '= - 0.582 / 10 1.7
h ’= 0.0989 m
h ’= -9.89 cm
f = 200 mm
h '= - 2.5 / 10 1.7
h ’= -0.425 m
h ’= 42.5 cm
The negative sign indicates that the image is real and inverted
Shadows blocking part of the light from the star.
A quick warning though this only works on planets either close to the star or planets that are very large.
Also to ensure that the shadows are planets the shadows have to move or orbit around the star. IE The shadow moves
C. motion can be defined as change in position.
According to Rubin, the desire to share your innermost thoughts and feelings is
c. intimacy
Hope this helps.
Which of what? Apologies please list the following so I can help.