(Application) You are asked to design a multiple lens system for the scenarios below. The lenses youhave available to you are:1. converging (f =7.5 cm)2. converging (f =15.0 cm)3. diverging (f = -15.0 cm)Think about the following scenarios:A) Lens 1 makes a real image i1, of which lens 2 makes a real image i2.B) Lens 1 makes a real image i1, of which diverging lens 2 makes a virtual image i2.C) Converging lens 1 makes a virtual image i1, of which converging lens 2 makes a virtual image i2.D) Converging lens 1 makes a virtual image i1, of which lens 2 makes a real image i2.For each scenario, decide:1) What type of lens is lens 1? Lens 2? Why? 2) Where should the object be placed with respect to the focal point to make the indicated type of image?3) Make a schematic drawing for Scenario D only. If you want extra practice, try to makeschematic drawings of the other three scenarios.

1) A) Lens converging, B) lens 1 converging

2) A and B object is at a greater distance than its focal

3)

Explanation:

For this exercise we will give some more important characteristics of the lenses.

we use the equation of the constructor

         1 / f = 1 / p + 1 / q

where f is the focal length and p and qlas the distance to the object and image, respectively.

We must also use how the lenses are combined

       1 / = 1 / f₁ + 1 / f₂

where feq is the equivalent focal length and it is assumed that the two lenses are in contact.

By applying the equation of the constructor you get some characteristics of the image

Converging lens

* Object farther than focal length real and inverted image

* Oject after focal virtual image and right

Diverging lens

* for all virtual image distance and right

with this information let's analyze the different configurations

1) let's do an analysis for each scenario

A) since lens 1 and lens 2 create real image they must be converging lenses

B) Lens 1 must be convergent and lens 2 can be of both types

C) Lens 1 can be convergent with the object that is at a shorter distance than at focal point (p <f₁). The same analysis diverges for lens 2, the object that is the image of the other lens is shorter than its focal distance p₂ <f₂

D) the object is at a shorter distance than the focal length p₁ <f₁, the object for this lens is at a greater distance than its focal point p₂> f₂

2) let's analyze for each scenario

A) for each lens the object is at a greater distance than its focal length p₁> f₁ and p₂> f₂

B) for lens 1 p₁> f₁

for lens 2

if it is convergent p₂ <f₂

and if it is divergent any distance is possible

C) lens 1 p₁ <f₁ and for lens 2 p₂ <f₂

D) lens 1 p₁ <f₁ and for lens 2 p₂> f₂

3) see attached

The distance with subscript 1 is measured with respect to line1 and the distance with subscript 2 is measured with respect to the lens with subscript 2

electromagnetic shielding is the practice of reducing the electromagnetic field in a space by blocking the field with barriers made of conductive or magnetic materials. shielding is typically applied to enclosures to isolate electrical devices from their surroundings, and to cables to isolate wires from the environment through which the cable runs. electromagnetic shielding that blocks radio frequency electromagnetic radiation is also known as rf shielding.

the shielding can reduce the coupling of radio waves, electromagnetic fields and electrostatic fields. a conductive enclosure used to block electrostatic fields is also known as a faraday cage. the amount of reduction depends very much upon the material used, its thickness, the size of the shielded volume and the frequency of the fields of interest and the size, shape and orientation of apertures in a shield to an incident electromagnetic

from wikipedia

total distance/total time

(85 + 82)/(5+3)

167/8

20 just over