Question 1

An object is placed 100 cm in front of a lens of focal length 20 cm.A second lens is placed 35 cm beyond the first,this second lens having a focal length of 8.0 cm.If the height of the object is 6.0 cm,what is the height of the final image?

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Question 2

It is desired to project the image of an object four times its actual size using a lens of focal length 20 cm.How far from the lens should the object be placed?

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Question 3

A thin double convex lens is to focus the image of an object onto a screen so that the image is life-sized.The lens surfaces have equal radii of curvature of 112 cm,and the refractive index of the lens material is 1.500.
(a)How far from the lens should the screen be placed?
(b)How far is the screen from the object?

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Question 4

A 4.0-cm tall object is placed 60 cm away from a converging lens of focal length 30 cm.What are the nature and location of the image? The image is

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Question 5

An object 1.80 cm tall is placed 100 cm in front of a diverging lens having a focal length of magnitude 25 cm.A converging lens with a focal length of magnitude 33.33 cm is placed 30 cm past the first lens.What is the lateral magnification of this system of lenses?

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First, we use the thin lens equation for the diverging lens:
1/f = 1/d_o - 1/d_i
where f is the focal length of the lens, d_o is the distance from the object to the lens, and d_i is the distance from the lens to the image.

Plugging in the given values, we get:
1/-25 = 1/100 - 1/d_i
Solving for d_i, we get:
d_i = -50 cm

This means the image formed by the diverging lens is virtual and located 50 cm behind the lens.

Now, we use the thin lens equation for the converging lens and the virtual image formed by the diverging lens:
1/f' = 1/d_i' - 1/d_o'
where f' is the focal length of the converging lens, d_i' is the distance from the image to the converging lens, and d_o' is the distance from the converging lens to the final image.

Plugging in the given values, we get:
1/33.33 = 1/-50 - 1/d_o'
Solving for d_o', we get:
d_o' = -30 cm

This means the final image is virtual and located 30 cm in front of the converging lens.

To find the lateral magnification, we use the formula:
m = -d_i'/d_o
where d_i' is the distance from the image to the converging lens and d_o is the distance from the object to the diverging lens.

Plugging in the given values, we get:
m = -(-50 cm)/100 cm * (-30 cm) = -0.40

Therefore, the lateral magnification of this system of lenses is -0.40 or C.

Question 6

In the figure,the image is viewed on a screen and is 13.5 mm tall.What is the focal length of the lens?

Accepted Answer

Using the thin lens equation, where f is the focal length, d_i is the object distance and d_o is the image distance:
1/f = 1/d_i + 1/d_o
As the image is viewed on a screen, the image distance can be considered essentially infinite. 
Therefore, 1/f = 1/d_i
Rearranging, d_i = f
Using similar triangles, we can write:
d_i / h_i = d_o / h_o
As the image height, h_i, is given as 13.5 mm and the object distance, d_i, is equal to the focal length, f, we can write:
f / 13.5 mm = d_o / h_o
Using the given values of h_o and f, we can solve for d_o:
d_o = f * h_o / h_i
Substituting the given values, we get:
d_o = f * 250 mm / 35 mm
Thus, d_o = 7.14f mm
As the image is formed behind the lens (inverted), the object distance, d_i, is negative. Therefore, f = -d_i.
Substituting, we get:
d_o = -7.14d_i
As d_i and d_o are both negative, we can drop the negative signs:
7.14d_i = d_o
Substituting the value of d_o, we get:
7.14d_i = f * 250 mm / 35 mm
Simplifying, we get:
d_i = f * 20.4 mm
Substituting the given value of h_i, we get:
f / 13.5 mm = 1 / (f * 20.4 mm)
Solving for f, we get:
f = 20.4 cm
Therefore, the focal length of the lens is +20.4 cm.

Question 7

In the figure,the radius of curvature of the curved part of the lens is 24.0 cm,and the refractive index of the lens material is 1.750.What is the focal length of the lens?

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Question 8

An object is placed 100 cm in front of a diverging lens with a focal length of magnitude 25 cm.A converging lens having a focal length of magnitude 33.33 cm is placed 30 cm past the first lens.Where is the final image formed?

Accepted Answer

Using the thin lens formula, 1/di = 1/f - 1/do, we can find the location of the image formed by the first diverging lens. Plugging in the values, we get 1/di = 1/-25 - 1/100, which simplifies to di = -20 cm (meaning the image is virtual and formed 20 cm in front of the lens).

Now we can use this virtual image as the object for the converging lens. The distance between the lenses is 30 cm, so we have do = -30 cm. Plugging in the values to the thin lens formula again, we get 1/di = 1/33.33 - 1/-30, which simplifies to di = 63.16 cm. This means the final image is real and formed 63.16 cm after the second lens, so the answer is D.

Question 9

When an object 1.15 cm tall is placed 12 cm from a lens,the lens produces an upright image of the object that is 5.75 cm tall.What is the focal length of the lens?

Accepted Answer

Using the formula:
m = -v/u = h'/h
where,
m = magnification
v = image distance
u = object distance
h' = height of image
h = height of object

Rearranging,
v = -m*u
and,
f = (u*v)/(u+v)

Substituting the given values,
m = 5.75/1.15 = 5
u = -12 cm
h' = 5.75 cm
h = 1.15 cm

Therefore,
v = -5*(-12) = 60 cm

and,
f = (-12*60)/(-12+60) = 15 cm

Hence, the correct choice is D) 15 cm.

Question 10

In the figure,the radius of curvature of the curved part of the lens is 35.0 cm,and the refractive index of the lens material is 1.620.What is the focal length of the lens?

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Question 11

A convex-concave thin lens is made with the radius of curvature of the convex surface being 25.0 cm and the concave surface 45.0 cm.If the glass used has index of refraction 1.500,what is the focal length of this lens?

Accepted Answer

The formula for the focal length of a thin lens is given by:
$$\frac{1}{f}=(n-1)\left(\frac{1}{R_1}-\frac{1}{R_2}\right)$$
where $n$ is the refractive index of the lens material, $R_1$ is the radius of curvature of the first surface (convex here) and $R_2$ is the radius of curvature of the second surface (concave here).
Substituting the given values in the above equation, we get:
$$\frac{1}{f}=(1.500-1)\left(\frac{1}{0.25}-\frac{1}{-0.45}\right)$$
Solving the above equation, we get $f=113$ cm. Therefore, the correct choice is C.

Question 12

A thin converging lens is found to form an image of a distant building 24 cm from the lens.If an insect is now placed 16 cm from this lens,how far FROM THE INSECT will its image be formed?

Accepted Answer

Using the lens formula, 1/f = 1/u + 1/v, where f is the focal length of the lens, u is the distance of the object from the lens, and v is the distance of the image from the lens.
Given u = 16 cm, v needs to be found. We know that the focal length of a thin lens is given by f = 1/(1/f1 + 1/f2), where f1 and f2 are the focal lengths of the two surfaces of the lens. For a converging lens, f1 and f2 are both positive. In this case, the lens is thin, so we can assume that f1 and f2 are approximately equal. Therefore, f ≈ f1/2 = f/2 → f ≈ 2f, or f ≈ 24 cm (from the given information).
Substituting the given values into the lens formula, 1/24 = 1/16 + 1/v → v = 48 cm.
Therefore, the image of the insect will be formed 48 cm from the lens on the opposite side. To find the distance from the insect to the image, we can use the formula v - u = (24 - 16) + (48 - v) → v - (48 - v) = 8 → v = 28 cm. Therefore, the distance from the insect to its image will be 28 - 16 = 12 cm. 
Hence the answer is option C.

Question 13

A 4.0-cm tall object is placed 50.0 cm from a diverging lens having a focal length of magnitude 25.0 cm.What are the nature and location of the image? The image is

Accepted Answer

Diverging lenses always produce virtual, upright, and smaller images than the object. The image location can be found using the lens equation 1/f = 1/d_o + 1/d_i, where f is the focal length, d_o is the object distance, and d_i is the image distance. Plugging in the values (with f = -25 cm for a diverging lens and d_o = 50 cm), we find d_i to be positive, indicating the image is on the same side as the object, and the calculation of image height using the magnification equation confirms the size.

Question 14

A lens is made with a focal length of -40 cm using a material with index of refraction 1.50.A second lens is made with the SAME GEOMETRY as the first lens,but using a material having refractive index of 2.00.What is the focal length of the second lens?

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Question 15

A double-convex thin lens is made of glass with an index of refraction of 1.52.The radii of curvature of the faces of the lens are 60 cm and 72 cm.What is the focal length of the lens?

Accepted Answer

Using the lens maker's formula (1/f = (n-1)(1/R1 - 1/R2)), where n is the index of refraction and R1 and R2 are the radii of curvature of the two faces, we can find the focal length to be 63 cm.

Question 16

An optical system comprises in turn,from left to right: an observer,a lens of focal length +30 cm,an erect object 20 mm high,and a convex mirror of radius 80 cm.The object is between the lens and the mirror,20 cm from the lens and 50 cm from the mirror.The observer views the image that is formed first by reflection and then by refraction.What is the position of the final image,measured from the mirror?

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Question 17

A double-concave lens has equal radii of curvature of  An object placed  from the lens forms a virtual image  from the lens.What is the index of refraction of the lens material?

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Question 18

A compound lens is made by joining the plane surfaces of two thin plano-convex lenses of different glasses.The radius of curvature of each convex surface is 80 cm.The indices of refraction of the two glasses are 1.50 and 1.60.What is the focal length of the compound lens?

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Question 19

An object 1.25 cm tall is placed 100 cm in front of a convex lens with a focal length of magnitude 50 cm.A concave lens with a focal length of magnitude 20 cm is placed 90 cm beyond the first lens.Where is the final image located?

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Question 20

An object is placed 10 cm from a convex lens of focal length 20 cm.What is the lateral magnification of the object?

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Quiz 34: Ray Optics