1) An object of height 1cm is 30 cm in front of a converging lens with a focal l

Question

1) An object of height 1cm is 30 cm in front of a converging lens with a focal length of 25 cm.

what is the magnification and how tall is the image?

2)How long(in ns) does it take light to travel 1millimeter in a piece of glass with index of refraction n=1.5?

3)A laser beam in glass (n=1.5) is incident on a liquid(n=1.2) at an angle of 20 degrees with respect to the normal.What is the refracted angle in the liquid?

4)1.90 mol of gas at a temperature of 500K fills a 15 m^3 container. What is the final pressure?

Explanation / Answer

1. To obtain this type of numerical information, it is necessary to use the Lens Equation and the Magnification Equation.

the lens equation::
1/f=1/d0+1/di

magnification equation:::
M=hi/h0=-di/d0

These two equations can be combined to yield information about the image distance and image height if the object distance, object height, and focal length are known.

solving the problem::::
h0=1cm
d0 =30cm
f=25cm
hi=?,di=?
so .04=.03+1/di
1/di=.01
di=100
hi/h0=di/d0
hi=100/30*1
hi=3.33cm(image height,magnification)

2.n=c/v

1.5=3.00*10^8/v

sov=.2ns

3.n1sin ?i = n2sin?r

1.5/1.2*sin20=sinr

0.34

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