Home Page Textbooks Physics Page 618: Lesson Check

Physics

1st Edition

Walker

ISBN: 9780133256925

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Page 618: Lesson Check

Exercise 35

Step 1

1 of 2

Since you want to focus sunlight you will need to use a convex lens because it focuses rays.

Result

2 of 2

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Exercise 36

Step 1

1 of 2

He is not right because the two equation look identical i.e.

$$
frac{1}{f}=frac{1}{d_0}+frac{1}{d_i}.
$$

Result

2 of 2

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Exercise 37

Step 1

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Mirrors form images from intersection of reflected ray (either their real parts or virtual extensions), while in the case of lenses refracted rays intersect to form an image. If the real rays intersect (which for the lens always happens on the opposite side from the object) the image is real and if virtual extensions of refracted rays intersect the image is virtual (which happens on the same side of the lens as the object is).

Result

2 of 2

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Exercise 38

Step 1

1 of 2

You should move the object closer to the lens because then because then the ray that goes from top of the object through the center of the lens will intercept the virtual extension of the refracted ray that goes through the focus higher and thus make the higher image.

Result

2 of 2

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Exercise 39

Step 1

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a) It moves closer to the lens because the ray going through the center of the lens will have smaller inclination and so it will intersect with the refracted ray going through the second focus closer.

b) I would be smaller and the explanation is the same as in a). Namely, if it intersects with it closer it will also intersect lower and make a smaller image.

Result

2 of 2

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Exercise 40

Step 1

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Convex lens is shaped so that it gathers all light rays that enter parallel to its central axis at a single point on the opposite side of the lens. Ray of light entering a convex lens converge at its focal point. When paper ignites, parallel rays from sunlight are focuses to a point on the paper. This is point where all light rays converges, and, hence, the focal point of the lens is $boxed{f = 26;text{cm}}$ .

Result

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$f = 26$ cm

Exercise 41

Step 1

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$textbf{Given values:}$

$$
begin{align*}
d_o &= 32 text{ cm} \
d_i &= 17 text{ cm}
end{align*}
$$

The focal length of the lens can be calculated by applying the thin-lens equation to the problem :

$$
{dfrac{1}{d_o} + dfrac{1}{d_i} = dfrac{1}{f}}
$$

$$
{f = dfrac{1}{dfrac{1}{d_o} + dfrac{1}{d_i}}}
$$

$$
{f = dfrac{1}{dfrac{1}{32 text{ cm}} + dfrac{1}{17 text{ cm}}}}
$$

$$
{boxed{f = 11.1 text{ cm}}}
$$

Result

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$$
{f = 11.1 text{ cm}}
$$

Exercise 42

Step 1

1 of 2

$textbf{Given values:}$

$$
begin{align*}
d_o &= 29 text{ cm} \
f &= 15 text{ cm}
end{align*}
$$

The image distance of the object can be calculated by applying the thin-lens equation to the problem :

$$
{dfrac{1}{d_o} + dfrac{1}{d_i} = dfrac{1}{f}}
$$

$$
{d_i = dfrac{1}{dfrac{1}{f} – dfrac{1}{d_o}}}
$$

$$
{d_i = dfrac{1}{dfrac{1}{15 text{ cm}} – dfrac{1}{29 text{ cm}}}}
$$

$$
{boxed{d_i = 31.07 text{ cm}}}
$$

Result

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$$
{d_i = 31.07 text{ cm}}
$$

Exercise 43

Step 1

1 of 2

$textbf{Given values:}$

$$
begin{align*}
d_i &= – 12 text{ cm} \
f &= – 23 text{ cm}
end{align*}
$$

The distance of the object from the focal point can be calculated by applying the thin-lens equation to the problem :

$$
{dfrac{1}{d_o} + dfrac{1}{d_i} = dfrac{1}{f}}
$$

$$
{d_o = dfrac{1}{dfrac{1}{f} – dfrac{1}{d_i}}}
$$

$$
{d_o = dfrac{1}{dfrac{1}{- 23 text{ cm}} + dfrac{1}{12 text{ cm}}}}
$$

$$
{boxed{d_o = 25.1 text{ cm}}}
$$

Result

2 of 2

$$
{d_o = 25.1 text{ cm}}
$$

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Page 618: Lesson Check

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