A 2.0-cm-tall candle flame is 2.0 m from a wall. You take place to havea lens via focal length of 32 cm.

You are watching: How many places can you put the lens to form a well-focused image of the candle flame on the wall?

A. How many kind of places deserve to you put the lens to form a well-concentrated imageof the candle on the wall?

B. For each position, what is the height and orientation of theimage?

Answer

A.

Take the difference between the object and image distance as 2m.

eginalignedv-u &=2 mathrm~m \v &=2 mathrm~m+uendalignedThe elevation of the object is 2.0 cm.

The expression for lens equation is as follows:

frac1f=frac1u+frac1v

The object is put on the left side of the lens. Thus, the above equation is recomposed as follows:

eginaligned frac1f &=frac1-u+frac1v \ &=frac1v-frac1u endaligned

Substitute 2m+u for v.

eginaligned&=fracu(2 m+u) u-frac(2 m+u)(2 m+u) u \&=fracu-2 m-u(2 m+u) u \&=frac-2 m(2 m+u) uendalignedAobtain, substitute 32cm for f.

eginalignedfrac132 mathrm~cmleft(frac10^-2 mathrm~m1 mathrm~cm ight) &=frac-2 mathrm~m(2 mathrm~m+u) u \frac10.32 mathrm~m &=frac-2 mathrm~m(2 mathrm~m+u) u \(2 mathrm~m+u) u &=-2 mathrm~m(0.32 mathrm~m) \u^2+2 u+0.64 &=0endalignedSolve the quadratic equation.

u=-0.4,-1.6 mathrm~m

Hence, the lens have the right to be inserted at 40 cm and also 160 cm to create a well-focused photo of the candle flame on the wall.

Therefore, The lens have the right to be inserted at two places.

B.

When the distance between the candle and also lens was 160 mathrm~cm. The expression for lens equation is as follows:

frac1f=frac1u+frac1v

Substitute 32 mathrm~cm for f and 160 mathrm~cm for u.

eginaligned frac132 mathrm~cm &=frac1160 mathrm~cm+frac1v \ frac1v &=frac132 mathrm~cm-frac1160 mathrm~cm \ &=frac140 mathrm~cm \ v &=40 mathrm~cm endalignedThe expression for magnification of the lens is as follows:

M=frac-vu

Substitute 40 mathrm~cm for v and also 160 mathrm~cm for u.

eginaligned M &=frac-40 mathrm~cm160 mathrm~cm \ &=-0.25 mathrm~cm endaligned

Anvarious other expression for magnification is as follows:

M=frach_ih_o

Compare the above 2 expressions and also rearvariety the expression for h_i as follows:

frac-vu=frach_mathrmih_mathrmo

h_mathrmi=frac-(v)u h_6

Substitute 40 mathrm~cm for v, 2 mathrm~cm for h_0, and 160 mathrm~cm for u.

h_i=frac-(40 mathrm~cm)160 mathrm~cm(2 mathrm~cm)

=-0.5 mathrm~cm

When the distance between the candle and also lens was 40 mathrm~cm.

The expression for lens equation is as follows:

M=frac-vu

Substitute 160 mathrm~cm for v and 40 mathrm~cm for u.

M=frac-160 mathrm~cm40 mathrm~cm

=-4 mathrm~cm

Another expression for magnification is as follows:

M=frach_ih_o

Compare the above two expressions and also rearselection the expression for h_i as follows:

frac-vu=frach_mathrmih_mathrme

h=frac-(v)u h_ㅇ

Substitute 160 mathrm~cm for $v, 2 mathrm~cm for h_0, and also 40 mathrm~cm for u.

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h_i=frac-(160 mathrm~cm)40 mathrm~cm(2 mathrm~cm)

=-8 mathrm~cm

Because of this, the height of the picture as soon as the distance between the candle and also lens are 160 mathrm~cm and also 40 mathrm~cm are -0.5 mathrm~cm and -8 mathrm~cm, respectively. In both cases, the magnification is negative, given that both are inverted.