Hi. This Physics lab report called a Simulation - Ray Optics lab. I need an expe

Question

Hi.

This Physics lab report called a Simulation - Ray Optics lab. I need an expert to answer all the questions and the measurement with very accurate results. I will attach the file with all the explanation and detailed information on it.

Thank you

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Explanation / Answer

Answer:-

Q-1 ) - Image found behind a lens is real image that will be Uprighted.

Q-2 ) -As the radius of curvature of the lens increases, the focal point of that lens becomes further away from that lens.

The focal length is always greater than the radisu oc curvature so if we increase the radius of curvature. the focal point of that lens

Q-3) As the refractive index of the lens increases, the focal point of that lens becomes closer to that lens.

Refractive index is inversely proportional to the focal length so if we increase the refractive index of the lens the focal point comes closer to that lens.

Q-4 ) What advantage does a larger lens have over a smaller lens (all other characteristics being equal)?

Ans- Advantage of larger lens over smaller lens as below,

1- With the same focal length a larger image create a brighter image than smaller image.

2- It gathers a more light and smaller f ratio than a smaller lens.

Q-5 What was the focal distance (f) when the radius of curvature was .70 and index of refraction was 1.8?

Ans- Given Data:- R = 0.7 m and n =1.8 .

Use a formula 1/f = (n-1) 2/R

1/f = (1.8-1) 2/0.7

f = 0.437 m

Q-6 ) Calculate the radius of curvature of a lens with a focal distance of 40. cm and an index of 1.2.

Ans- Given Data f= 40 cm =0.4 m

n =1.2

use formula

1/f = (n-1) 2/R

1/ 0.4 = (1.2-1) 2/R

R =0.16 m

Q-7) An object placed 35cm away from a lens projects a real image .55m behind the lens. What is this lens’ focal distance?

Ans- do = 35 cm = 0.35 m di= 0.55 m f= ?

use formual

1/f = 1/do + 1/di

1/f = 1/0.35 + 1/(-0.55)

f = 0.962 m

Q- 8) What is the lens’ magnification?

Ans- A magnification can be define as the process of enlarging anything by appearance not in physical. and the enlargement of the any object calcualted by the numbers that is called a magnifications.

Q-9 ) An object 20. 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.

Ans - Given Data ho = 1 m , m = -2

use formula m = hi/ ho

-2 =hi /1

hi = -2 m

hi = - 200 cm

it is on the right side of the lens.

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