Question
1.
Light leaves a piece of glass (n = 1.4) at an angle of 30° from the normal. It enters air (n = 1). What is the angle of refraction on the air?
54°
34°
24°
44°
2. Which statements are true regarding refraction? Check all that apply.
Light bends towards the normal when entering a lower index of refraction
Light bends towards the normal when entering a higher index of refraction
Light bends away from the normal when entering a lower index of refraction
Light bends away from the normal when entering a higher index of refraction
3. The focal distance for a convex lens is 80 cm. What kind of image is produced when the object distance is between 80 cm and 160 cm?
upright smaller
inverted and smaller
upright and larger
inverted and larger
4. The magnification equation is:
m = hi/ha
m=si/so
m=ho/hi
m=-so/si
5. What kind of images are formed when an object is brought inside the focal point of a convex lens?
upright and larger
inverted and larger
upright and smaller
inverted and smaller
6. A diverging lens has a focal length of –10 cm. Suppose an object is placed 15 cm in front of the lens. a) What is the image distance? b) Is the image real or virtual?
a) 6 cm b) real
a) -10 cm b) real
a) -8 cm b) virtual
a) -6 cm b) virtual
7. Which rules of ray tracing apply to converging lenses? Check all that apply.
A ray that passes through the center of the lens and then parallel to the optical axis on the other side of the lens.
A ray that comes through the focal point on the way to the lens will refract and travel parallel to the principal axis on the other side.
A ray that comes into the lens parallel to the principal axis will refract through the lens and travel through the focal point.
A ray that passes through the center (i.e. the principal axis) will continue traveling in that same direction.
8.
A lens is:
a transparent object that refracts light rays and causes the formation of images.
a translucent object that refracts light rays and causes the formation of images.
a transparent object that reflects light rays and causes the formation of images.
a translucent object that reflects light rays and causes the formation of images.
9.
What is the critical angle for light bouncing off a fish in water of nw = 1.35 and the air above having an index of na = 1.00?
50.0 degrees
45.0 degrees
47.8 degrees
48.8 degrees
10.
A converging lens has a focal length of 40 cm. If an object is placed 50 cm in front of the image, where will the image be formed?
200 cm to the right side of the lens
250 cm to the right side of the lens
200 cm to the left side of the lens
150 cm to the left side of the lens
Light leaves a piece of glass (n = 1.4) at an angle of 30° from the normal. It enters air (n = 1). What is the angle of refraction on the air?
54°
34°
24°
44°
2. Which statements are true regarding refraction? Check all that apply.
Light bends towards the normal when entering a lower index of refraction
Light bends towards the normal when entering a higher index of refraction
Light bends away from the normal when entering a lower index of refraction
Light bends away from the normal when entering a higher index of refraction
3. The focal distance for a convex lens is 80 cm. What kind of image is produced when the object distance is between 80 cm and 160 cm?
upright smaller
inverted and smaller
upright and larger
inverted and larger
4. The magnification equation is:
m = hi/ha
m=si/so
m=ho/hi
m=-so/si
5. What kind of images are formed when an object is brought inside the focal point of a convex lens?
upright and larger
inverted and larger
upright and smaller
inverted and smaller
6. A diverging lens has a focal length of –10 cm. Suppose an object is placed 15 cm in front of the lens. a) What is the image distance? b) Is the image real or virtual?
a) 6 cm b) real
a) -10 cm b) real
a) -8 cm b) virtual
a) -6 cm b) virtual
7. Which rules of ray tracing apply to converging lenses? Check all that apply.
A ray that passes through the center of the lens and then parallel to the optical axis on the other side of the lens.
A ray that comes through the focal point on the way to the lens will refract and travel parallel to the principal axis on the other side.
A ray that comes into the lens parallel to the principal axis will refract through the lens and travel through the focal point.
A ray that passes through the center (i.e. the principal axis) will continue traveling in that same direction.
8.
A lens is:
a transparent object that refracts light rays and causes the formation of images.
a translucent object that refracts light rays and causes the formation of images.
a transparent object that reflects light rays and causes the formation of images.
a translucent object that reflects light rays and causes the formation of images.
9.
What is the critical angle for light bouncing off a fish in water of nw = 1.35 and the air above having an index of na = 1.00?
50.0 degrees
45.0 degrees
47.8 degrees
48.8 degrees
10.
A converging lens has a focal length of 40 cm. If an object is placed 50 cm in front of the image, where will the image be formed?
200 cm to the right side of the lens
250 cm to the right side of the lens
200 cm to the left side of the lens
150 cm to the left side of the lens
Answers
There are no human answers yet.
Answered by
GPT-4o mini
Let's go through each question one at a time.
1. **Light leaves a piece of glass (n = 1.4) at an angle of 30° from the normal. It enters air (n = 1). What is the angle of refraction in the air?**
We use Snell's Law:
\[
n_1 \sin(\theta_1) = n_2 \sin(\theta_2)
\]
where \(n_1 = 1.4\), \(\theta_1 = 30°\), \(n_2 = 1\) and \(\theta_2\) is the angle of refraction.
\[
1.4 \sin(30°) = 1 \sin(\theta_2)
\]
\[
1.4 \times 0.5 = \sin(\theta_2)
\]
\[
0.7 = \sin(\theta_2)
\]
Therefore, \(\theta_2 = \sin^{-1}(0.7) \approx 44.4°\).
**Answer: 44°**
2. **Which statements are true regarding refraction?** (Check all that apply)
- Light bends towards the normal when entering a lower index of refraction: **False**
- Light bends towards the normal when entering a higher index of refraction: **True**
- Light bends away from the normal when entering a lower index of refraction: **True**
- Light bends away from the normal when entering a higher index of refraction: **False**
**True statements:**
- Light bends towards the normal when entering a higher index of refraction
- Light bends away from the normal when entering a lower index of refraction
3. **What kind of image is produced when the object distance is between 80 cm and 160 cm (focal distance is 80 cm)?**
For a convex lens, when the object distance \(d_o\) > focal length, the image is real and inverted. As \(d_o\) increases from the focal length to the double focal length, the image becomes smaller.
**Answer: inverted and smaller**
4. **The magnification equation is:**
The magnification equation can be given by:
- \( m = \frac{h_i}{h_a} \)
- \( m = \frac{s_i}{s_o} \)
- \( m = \frac{h_o}{h_i} \) (generally not used in practical lens calculations)
- \( m = -\frac{s_o}{s_i} \)
The relevant magnification formulas are typically:
\[
m = \frac{h_i}{h_o} = \frac{s_i}{s_o} = -\frac{s_o}{s_i}
\]
**Answer: m = hi/ha, m=si/so, m=-so/si** (m = ho/hi is not typically correct).
5. **What kind of images are formed when an object is brought inside the focal point of a convex lens?**
When the object is inside the focal point, the image is virtual, upright, and larger.
**Answer: upright and larger**
6. **A diverging lens has a focal length of -10 cm. The object is placed 15 cm in front of the lens. a) What is the image distance? b) Is the image real or virtual?**
Using the lens formula:
\[
\frac{1}{f} = \frac{1}{s_o} + \frac{1}{s_i}
\]
\[
\frac{1}{-10} = \frac{1}{15} + \frac{1}{s_i}
\]
Solving for \( s_i \):
\[
-0.1 = \frac{1}{15} + \frac{1}{s_i}
\]
\[
\frac{1}{s_i} = -0.1 - \frac{1}{15} \text{ which gives } s_i = -6 cm
\]
Since \(s_i\) is negative, the image is virtual.
**Answer: a) -6 cm b) virtual**
7. **Which rules of ray tracing apply to converging lenses?** (Check all that apply)
- A ray that passes through the center of the lens and then parallel to the optical axis on the other side of the lens: **False**
- A ray that comes through the focal point on the way to the lens will refract and travel parallel to the principal axis on the other side: **True**
- A ray that comes into the lens parallel to the principal axis will refract through the lens and travel through the focal point: **True**
- A ray that passes through the center (i.e. the principal axis) will continue traveling in that same direction: **True**
**True statements:**
- A ray that comes through the focal point on the way to the lens will refract and travel parallel to the principal axis on the other side.
- A ray that comes into the lens parallel to the principal axis will refract through the lens and travel through the focal point.
- A ray that passes through the center (i.e. the principal axis) will continue traveling in that same direction.
8. **A lens is:**
A lens is defined as a transparent object that refracts light rays and causes the formation of images.
**Answer: a transparent object that refracts light rays and causes the formation of images.**
9. **What is the critical angle for light bouncing off a fish in water of nw = 1.35 and the air above having an index of na = 1.00?**
The critical angle \(θ_c\) can be found using Snell's Law:
\[
n_1 \sin(θ_c) = n_2 \sin(90°)
\]
\[
1.35 \sin(θ_c) = 1.00
\]
\[
\sin(θ_c) = \frac{1.00}{1.35} \approx 0.7407
\]
\[
θ_c = \sin^{-1}(0.7407) \approx 47.8°.
\]
**Answer: 47.8 degrees**
10. **A converging lens has a focal length of 40 cm. If an object is placed 50 cm in front of the lens, where will the image be formed?**
Using the lens formula:
\[
\frac{1}{f} = \frac{1}{s_o} + \frac{1}{s_i}
\]
\[
\frac{1}{40} = \frac{1}{50} + \frac{1}{s_i}
\]
Solving for \( s_i \):
\[
\frac{1}{s_i} = \frac{1}{40} - \frac{1}{50}
\]
Find a common denominator (200):
\[
\frac{5}{200} - \frac{4}{200} = \frac{1}{200}
\]
So, \( s_i = 200 cm \).
Since the image distance is positive, the image is formed on the opposite side of the lens.
**Answer: 200 cm to the right side of the lens**
1. **Light leaves a piece of glass (n = 1.4) at an angle of 30° from the normal. It enters air (n = 1). What is the angle of refraction in the air?**
We use Snell's Law:
\[
n_1 \sin(\theta_1) = n_2 \sin(\theta_2)
\]
where \(n_1 = 1.4\), \(\theta_1 = 30°\), \(n_2 = 1\) and \(\theta_2\) is the angle of refraction.
\[
1.4 \sin(30°) = 1 \sin(\theta_2)
\]
\[
1.4 \times 0.5 = \sin(\theta_2)
\]
\[
0.7 = \sin(\theta_2)
\]
Therefore, \(\theta_2 = \sin^{-1}(0.7) \approx 44.4°\).
**Answer: 44°**
2. **Which statements are true regarding refraction?** (Check all that apply)
- Light bends towards the normal when entering a lower index of refraction: **False**
- Light bends towards the normal when entering a higher index of refraction: **True**
- Light bends away from the normal when entering a lower index of refraction: **True**
- Light bends away from the normal when entering a higher index of refraction: **False**
**True statements:**
- Light bends towards the normal when entering a higher index of refraction
- Light bends away from the normal when entering a lower index of refraction
3. **What kind of image is produced when the object distance is between 80 cm and 160 cm (focal distance is 80 cm)?**
For a convex lens, when the object distance \(d_o\) > focal length, the image is real and inverted. As \(d_o\) increases from the focal length to the double focal length, the image becomes smaller.
**Answer: inverted and smaller**
4. **The magnification equation is:**
The magnification equation can be given by:
- \( m = \frac{h_i}{h_a} \)
- \( m = \frac{s_i}{s_o} \)
- \( m = \frac{h_o}{h_i} \) (generally not used in practical lens calculations)
- \( m = -\frac{s_o}{s_i} \)
The relevant magnification formulas are typically:
\[
m = \frac{h_i}{h_o} = \frac{s_i}{s_o} = -\frac{s_o}{s_i}
\]
**Answer: m = hi/ha, m=si/so, m=-so/si** (m = ho/hi is not typically correct).
5. **What kind of images are formed when an object is brought inside the focal point of a convex lens?**
When the object is inside the focal point, the image is virtual, upright, and larger.
**Answer: upright and larger**
6. **A diverging lens has a focal length of -10 cm. The object is placed 15 cm in front of the lens. a) What is the image distance? b) Is the image real or virtual?**
Using the lens formula:
\[
\frac{1}{f} = \frac{1}{s_o} + \frac{1}{s_i}
\]
\[
\frac{1}{-10} = \frac{1}{15} + \frac{1}{s_i}
\]
Solving for \( s_i \):
\[
-0.1 = \frac{1}{15} + \frac{1}{s_i}
\]
\[
\frac{1}{s_i} = -0.1 - \frac{1}{15} \text{ which gives } s_i = -6 cm
\]
Since \(s_i\) is negative, the image is virtual.
**Answer: a) -6 cm b) virtual**
7. **Which rules of ray tracing apply to converging lenses?** (Check all that apply)
- A ray that passes through the center of the lens and then parallel to the optical axis on the other side of the lens: **False**
- A ray that comes through the focal point on the way to the lens will refract and travel parallel to the principal axis on the other side: **True**
- A ray that comes into the lens parallel to the principal axis will refract through the lens and travel through the focal point: **True**
- A ray that passes through the center (i.e. the principal axis) will continue traveling in that same direction: **True**
**True statements:**
- A ray that comes through the focal point on the way to the lens will refract and travel parallel to the principal axis on the other side.
- A ray that comes into the lens parallel to the principal axis will refract through the lens and travel through the focal point.
- A ray that passes through the center (i.e. the principal axis) will continue traveling in that same direction.
8. **A lens is:**
A lens is defined as a transparent object that refracts light rays and causes the formation of images.
**Answer: a transparent object that refracts light rays and causes the formation of images.**
9. **What is the critical angle for light bouncing off a fish in water of nw = 1.35 and the air above having an index of na = 1.00?**
The critical angle \(θ_c\) can be found using Snell's Law:
\[
n_1 \sin(θ_c) = n_2 \sin(90°)
\]
\[
1.35 \sin(θ_c) = 1.00
\]
\[
\sin(θ_c) = \frac{1.00}{1.35} \approx 0.7407
\]
\[
θ_c = \sin^{-1}(0.7407) \approx 47.8°.
\]
**Answer: 47.8 degrees**
10. **A converging lens has a focal length of 40 cm. If an object is placed 50 cm in front of the lens, where will the image be formed?**
Using the lens formula:
\[
\frac{1}{f} = \frac{1}{s_o} + \frac{1}{s_i}
\]
\[
\frac{1}{40} = \frac{1}{50} + \frac{1}{s_i}
\]
Solving for \( s_i \):
\[
\frac{1}{s_i} = \frac{1}{40} - \frac{1}{50}
\]
Find a common denominator (200):
\[
\frac{5}{200} - \frac{4}{200} = \frac{1}{200}
\]
So, \( s_i = 200 cm \).
Since the image distance is positive, the image is formed on the opposite side of the lens.
**Answer: 200 cm to the right side of the lens**
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