I would appreciate an explanation or drawing with the answer. https://phet.color
ID: 1270649 • Letter: I
Question
I would appreciate an explanation or drawing with the answer.
https://phet.colorado.edu/sims/geometric-optics/geometric-optics_en.html
6. Repeat step 5 with a different prescription. The refractive index of the lens is 1.6 and radius of curvature of 0.8. Remember to find out the focal length first.
f = ___________
Distance of the object (p)
Distance of the image (q)
Magnification (m)
Nature of the image (Real or Virtual/upright or inverted)
190 cm
120 cm
90 cm
60 cm
30 cm
7. Determine the values of refractive index and radius of curvatures of different combinations to find the focal lengths given below
Refractive index (n)
Radius of curvature (R)
Focal length (f)
8. Set the refractive index (n) to 1.5 and the radius of curvature (R) to 0.6 m. Position the object so that the image is formed 100 cm to the right of the lens (use ruler). Is the image enlarged or reduced in size?
Size: _______
What is the object position (use ruler) ? ___________
What is the magnification? __________
Now keeping the object and ruler fixed (between lens center and image position at 100 cm), adjust the LENS position (move it left or right) so that the image is again formed in the same place as before (i.e. at the same mark on the ruler as before).
What are the new object and image positions? Use the ruler to measure. _____
Is the image enlarged or reduced in size? ______
Do you observe a relationship between the new object and image position vs the old object and image positions? What is this relation? Explain why this is so.
Conclusion questions and Calculations:
1. Images found on the opposite side (as the object) of a lens are real / virtual images that will be upright / inverted.
2. As the radius of curvature of the lens increases, the focal point of that lens becomes closer to / further away from that lens.
3. As the refractive index of the lens increases, the focal point of that lens becomes closer to / further away from that lens.
4. What advantage does a larger lens have over a smaller lens (all other characteristics being equal)?
5. What was the focal distance (f ) when the radius of curvature was 0.8 and index of refraction was 1.6? _____________
6. Calculate the radius of curvature of a lens with a focal distance of 40.0 cm and an index of 1.2. __________________
7. An object placed 3cm away from a lens projects a real image 0.6m behind the lens. What is this lens focal distance? _____________________
8. What is the lens magnification? ____________________
9. An object 20.0 cm to the left of a convex lens is 1.0 m in height. What is the height and location of its image if the lens has a magnification of -2.0?
_____________ m and _______________cm on the left / right side of the lens.
What is the focal length of the lens? ____________
10. Imagine you are nearsighted. In order to correct your vision, you need glasses with a converging / diverging lens. Explain your answer.
Distance of the object (p)
Distance of the image (q)
Magnification (m)
Nature of the image (Real or Virtual/upright or inverted)
190 cm
120 cm
90 cm
60 cm
30 cm
Explanation / Answer
6. Repeat step 5 with a different prescription. The refractive index of the lens is 1.6 and radius of curvature of 0.8. Remember to find out the focal length first.
F = 0.4/(0.6) = 66 cm
Distance of the object (p)
Distance of the image (q)
Magnification (m)
Nature of the image (Real or Virtual/upright or inverted)
190 cm
120 cm
90 cm
60 cm
30 cm
7. Determine the values of refractive index and radius of curvatures of different combinations to find the focal lengths given below (Please give the focal lengths as said in the question? still i will try to put few combination)
Refractive index (n)
Radius of curvature (R)
Focal length (f)
8. Set the refractive index (n) to 1.5 and the radius of curvature (R) to 0.6 m. Position the object so that the image is formed 100 cm to the right of the lens (use ruler). Is the image enlarged or reduced in size?
Size: reduced in size
What is the object position (use ruler) ? 148cm
What is the magnification? -0.675
Now keeping the object and ruler fixed (between lens center and image position at 100 cm), adjust the LENS position (move it left or right) so that the image is again formed in the same place as before (i.e. at the same mark on the ruler as before).
What are the new object and image positions? Use the ruler to measure.
object position : 100 cm, image position: 148
Is the image enlarged or reduced in size? image is enlarged
Do you observe a relationship between the new object and image position vs the old object and image positions? What is this relation? Explain why this is so.
New object and new image position has been respectively interchanged with old image and old object position.
Conclusion questions and Calculations:
1. Images found on the opposite side (as the object) of a lens are real / virtualimages that will be upright / inverted.
Images found on the opposite side (as the object) of a lens are real / virtual images that will be inverted /upright
2. As the radius of curvature of the lens increases, the focal point of that lensbecomes closer to / further away from that lens.
As the radius of curvature of the lens increases, the focal point of that lens becomes further away from that lens.
3. As the refractive index of the lens increases, the focal point of that lens becomes closer to / further away from that lens.
As the refractive index of the lens increases, the focal point of that lens becomes closer to from that lens.
4. What advantage does a larger lens have over a smaller lens (all other characteristics being equal)?
A larger lens refracts more light thus increasing intensity of image which brings clarity in image and quality of image increases.
5. What was the focal distance (f ) when the radius of curvature was 0.8 and index of refraction was 1.6?
66cm
6. Calculate the radius of curvature of a lens with a focal distance of 40.0 cm and an index of 1.2.
r = 2*40*0.2 = 16 cm
7. An object placed 3cm away from a lens projects a real image 0.6m behind the lens. What is this lens focal distance?
f = oi/(o+i) = 60*3/(60+3) = 2.857 cm
8. What is the lens magnification?
m = -60/3 = -20
9. An object 20.0 cm to the left of a convex lens is 1.0 m in height. What is the height and location of its image if the lens has a magnification of -2.0?
2 m and 40 cm on the right side of the lens.
What is the focal length of the lens?
13.33 cm
10. Imagine you are nearsighted. In order to correct your vision, you need glasses with a converging / diverging lens. Explain your answer.
Connverging lens.
Since near sightedness will occurs because rays are not getting sufficiently coverged through eye lens to form real images on retina. Hence to ease convergence of rays we use converging lens.
Distance of the object (p)
Distance of the image (q)
Magnification (m)
Nature of the image (Real or Virtual/upright or inverted)
190 cm
103 cm -0.54 Real, inverted120 cm
150cm -1.25 Real, Inverted90 cm
256 cm -2.84 Real, inverted60 cm
-660 cm 11 Inverted, upright30 cm
-55 cm 1.833 Inverted, upright