A camera with a 100mm lens can be used to focus objects from 6pm to infinity onto screen. how much must the lens be moved to focus on the extremities of this range

i = f = 0.1 m until the lens moves towards the screen 0.1 m

Explanation:

For this exercise let's use the constructor equation

         1 / f = 1 / o + 1 / i

Where f is the focal distance, or the distance to the object and "i" the distance to the image

It indicates that the focal distance is 100 mm (f = 100 mm), when an object is at infinity the image is formed at its focal length

     1 / f = 1 / inf + 1 / i = 1 / i

      i = f = 100 mm

At this point the screen is placed

For the shortest distance the lens has to move a little so the distance to the image is

        f = 100 mm = 0.1 m

        o = 6 pm = 6 10⁻¹² m

        i = 100 + x

        1 / f = 1 / o + 1 / 100+ x

        1 /( 0.10 + x) = 1 / f - 1 / o

        1 / 0.100+ x = 1/0.100 - 1/6 10-12

        1 / 0.100 + x = 10 - 10¹¹ = -10¹¹

        0.100 + x = -10⁻¹¹

        x = -10⁻¹¹ -10⁻¹

        x = -10⁻¹

        x = - 0.1 m

This negative distance indicates that the lens moves towards the screen 0.1 m

gpe = mass x gravity constant x height or… gpe = mgh

where… mass: measured in kilograms (kg)

gravity: gravity constant on earth = 10.00 m/s2

height: measured in meters (m)

gpe: measured in joules (j) or kilojoules (kj)

how do you solve gpe?

gravitational potential energy (gpe) is the energy an object has because of its position above the. ground.

gpe = mass x gravity constant x height or… gpe = mgh.

mass:

note: to earn full marks when solving science word problems, you must show your work:

formula.

example:

given values:

mass = 0.3 kg.

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