n = 1.7; r = 0.5 (unit is not given. so I take in m) .
from formula: 1/f = (n-1)*2/r
f = 0.357 m.
(a) s = 190 cm = 0.190 m
from lens formula: 1/f = 1/s + 1/s' (1)
image distance (s' )= - 0.40 m
magnification: m = -s'/s = 2.10 (2)
due to negative sign of image it will be virtual. since M is positive so it will be upright .
Note: Because unit of radius of curvature is not clear in question. so I am not solving other parts. You can calculate others parts from formula (1) and (2).
Nature of image can be determindes as: if s' is positive then real image and if negative then virtual image.
if m is positive then upright and if negative than inverted.
AaBbCD A. AaBb C 5. Repeat step 4 for a lens with different property. The refractive...
this is what they gave us
4. Use the focal length you found in (3), complete the table below and check your work in the simulation. Object Distance (p) Image Distance (9) Magnification Nature of the image (Real or (m) Virtual/upright or inverted) 190 cm 120 cm 90 cm 60 cm 30 cm Online Lab #9 3. Verify the focal length of the lens: a. Set the refractive index (n) to 1.5 and the radius of curvature (R) to 0.6...
1.) An object is placed in front of a diverging lens with a focal length of 17.7 cm. For each object distance, find the image distance and the magnification. Describe each image. (a) 35.4 cm location _____cm magnification _____ nature real virtual upright inverted (b) 17.7 cm location _____ cm magnification _____ nature real virtual upright inverted (c) 8.85 cm location _____ cm magnification _____ nature real virtual upright inverted 2.) An object is placed in front of a converging lens...
A diverging lens has a focal length of magnitude 21.2 cm. (a) Locate the images for each of the following object distances. 42.4 cm distance cm location ---Select---in front of the lensbehind the lens 21.2 cm distance cm location ---Select---in front of the lensbehind the lens 10.6 cm distance cm location ---Select---in front of the lensbehind the lens (b) Is the image for the object at distance 42.4 real or virtual? real virtual Is...
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A diverging lens has a focal length of magnitude 23.8 cm. (a) Locate the images for each of the following object distances. 47.6 cm distance cm location 23.8 cm distance cm location 11.9 cm distance cm location (b) Is the image for the object at distance 47.6 real or virtual? real virtual Is the image for the object at distance 23.8 real or virtual? real virtual Is the image for the object at distance 11.9 real or virtual? real virtual...
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A diverging lens has a focal length of magnitude 21.2 cm. (a) Locate the images for each of the following object distances. 42.4 cm distance cm location in front of A 21.2 cm distance location cm in front of 10.6 cm distance location in front of cm 4 (b) Is the image for the object at distance 42.4 real or virtual? o real o virtual Is the image for the object at distance 21.2 real or virtual? o real o...
I need 4 and 6 and show all work please
4) A 5 cm tall object is placed in front 20 cm. The object is located 20 cm from the spherical mirror of a spherical mirror with a radius of curvature of - a) What is the object distance? b) What is the image distance? c) What is the focal length? d) What is the size of the object? e) What is the size of the image? 1) What is...
A diverging lens has a focal length of magnitude 19.8 cm. (a) Locate the images for each of the following object distances 39.6 cm dstanice Your response daffers significandy from the correct answer: Reswork your soluation from the begining and check each step caruly,.om 19.8 cm location [n 9.9 ơn (b) Is the image for the object at distance 39.6 real or virtual? real e virtual is the image for the object at distance 19.8 real or virtual? O real...
a.) Determine the image distance for an object d0 = 6.500 cm
from a diverging lens of radius of curvature 5.200 cm and index of
refraction 1.750. (Express your answer as a positive quantity.)
b.)What is the magnification of the object? (Be sure to insert
the proper sign if needed.)
c.)what is the nature of the image
Determine the image distance for an object d 6.500 cm from a diverging lens of radius of curvature 5.200 cm and index of...