You have a thin converging lens whose focal length is 0.3m and a sheet of white paper on which to display a real image of a small tree that is 2m tall and 40m way.How far in back of the lens should you place the paper in order to get a sharp image of the tree? How tall is the image of the tree? Is it inverted or right-side up?

Short Answer

Expert verified

The paper must be placed at the distance of 0.302m. The height of image of the tree is 0.0015m. The image of the tree is inverted.

Step by step solution

01

Explain the given information

Consider that the focal length of a thin converging lens is 0.3m. A sheet is kept 40m away to display the real image of small tree of height 2m.

02

Give the Formula used.

Give the formula for magnification in terms of height and distance.

M=h2h1 ……(1)

M=d2d1 ……(2)

Give the lens equation.

1f=1d1+1d2 ……(3)

03

Step 3:  How far to place the paper in order to get a sharp image of the tree? How tall is the image of the tree? Is it inverted or right-side up?

Consider the magnification formula in terms of height from Equation (1),

M=h2h1

Consider the magnification formula in terms of distance from Equation (2),

M=d2d1

Compare the Equations (1) and (2),

h2h1=d2d1 ……(4)

Solve the Equation (3) for d2as follows,

d2=fd1d1-f ……(5)

Substitute the values f=0.3mand d1=40min Equation (5),

d2=0.3m40m40m-0.3m

d2=0.302m

Solve the Equation (4) for h2 as follows,

h2=-d2d1h1 ……(6)

Substitute the values d2=0.302m, d1=40mand h1=2m into the Equation (6).

h2=-0.302m40m2mh2=-0.015m

The negative sign represents that the height of the image is decreased.

The magnification of the image is,

M=-d2d1 ……(7)

Substitute the values d2=0.302m,d1=40minto the Equation (7)

M=-0.302m40mM=-0.008

Hence, the image is inverted.

Therefore, The paper must be placed at the distance of 0.302m. The height of image of the tree is 0.0015m. The image of the tree is inverted

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