Problem #1 (15 points) Part (a) of the figure below shows the eye of a person lo

Question

Problem #1 (15 points) Part (a) of the figure below shows the eye of a person looking through a lens so that the index of refraction of the medium that the incident light is moving through is ni -1.2 and the person's eye is in a medium having an index of refraction given by nR -1.8, which is also the index of refraction of the medium that the refracted light moves through into the eye of the person If the focal position of this lens is fa +360 cm, determine the focal position of the lens if the person moves their eye to the other side of the lens, as shown in part (b) of the figure. Note that here, the index of refraction of the medium that the incident light is moving through is now n 1.8 and the person's eye is in a medium having an index of refraction given by nR 1.2, which is also the index of refraction of the medium that the refracted light moves through into the eye of the person.

Explanation / Answer

From lens nakers formula we have

1/f =(nlens - n°)/n° *(1/R1 - 1/R2)

Focal length =360

Here nlens =1.2

n°=1.8

Put values in formula we get

1/360 =(1.2 - 1.8/1.8)*(1/R1 - 1/R2)

1/360 =(-1/3)*(1/R1-1/R2)

(1/R1 - 1/R2)=—3/360 =—1/120

In case 2, we have

Nlens=1.8

n°=1.2

R1 and R2 will interchange

So we get

1/f =(1.8-1.2/1.2)*(1/R2 - 1/R1)

1/f =(1/2)*(1/120)

1/f=1/240

f=240 cm

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