Question

An object is placed 66.0 cm from a screen.

(a) Where should a converging lens of focal length 7.5 cm be placed to form a clear image on the screen? (Give your answer to at least one decimal place.)

shorter distance	cm from the screen
farther distance	cm from the screen

(b) Find the magnification of the lens.

magnification if placed at the shorter distance
magnification if placed at the farther distance

Answer 1

converging lens of focal length f = 7.5 cm

An object is placed 66.0 cm from a screen.

From the thin lens equation,

1/p + 1/q= 1/? ,

and p + q = 66.0cm

so p =66.0cm -q

the image distance is found to be 1/(66 − q )+ 1/q = 1/? → 66q(66− q) = 1/? →

66? = −q² + 66q

q²-66q+66?=0

since f=7.5cm so plugging the value in the above formula

66(7.5cm) −q² + 66q=0

q² − 66q +495= 0.

which is a quadratic equation

so roots of a quadratic equation

x=q and b= -66 ,a=1 and c= 495

q=(66 ± ((66)²-((4)(1)(495))^1/2 ) /2 =  

q =33±24.37 =

q = 57.37cm or q = 8.63 cm  

shorter distance q1	= 8.628 cm from the screen
farther distance q2	= 57.37 cm from the screen

The magnification of the lens is

m₁ = -q₁/p_{1 =} -q₁/(66.0cm -q₁) = -8.628 /(66-8.628) = -8.628/57.372 = -0.15 cm

m2 = -q₂/p_{2 =} -q₂/(66.0cm -q₂) = -57.37 /(66-57.37) = -57.37/8.63 = -6.648 cm

magnification if placed at the shorter distance m1= -0.15 cm magnification if placed at the farther distance    m2= -6.648 cm

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