Answered: An object with a height 5cm lies d.= 2… | bartleby

Related questions

Question

3. An object with a height 5cm lies do= 2 x f in front of a lens with focal length of
f-(#N) m m. Find the location of image and the magnification.
a) 0.3N m;.5 b)2x0.N cm;-3/2 c)Nx1.5 cm;0.75 d)4x 0.00N m; -1
e) none of these is true

Expert Solution

Trending now

This is a popular solution!

Step by step

Solved in 2 steps with 1 images

SEE SOLUTION Check out a sample Q&A here

Blurred answer

Similar questions

12). A stamp collector wants to magnify images by 5(X) when the object is 6.2-cm away. What focal legt is needed? f- di - do =
2) A thin converging lens has a focal length of 4 cm. A 2 cm tall object is placed 9 cm to the left of the lens. | Carefully (i.e. using a ruler to measure) ray trace (draw the 3 principle rays) to locate the image of this object.
[12]
The lens-makers' equation can be written as 1 7- (22²-¹²/²-1/2) (R₁₂ = n₁ R₂ where n₂ represents the index of refraction of the lens material and n₁ is that of the medium surrounding the lens. (a) A certain lens has focal length 57.8 cm in the air and index of refraction 1.55. Find its focal length in water. (Give your answer in cm.) cm (b) A certain mirror has focal length 57.8 cm in the air. Find its focal length in water. (Give your answer in cm.) cm
3) (20 pts.) Compute the focal length of a biconcave thin lens immersed in water. The lens is made of flint glass, with index of refraction 1.66. The radii of the surfaces of the lens are 10 cm for the left surface and 20 cm for the right surface (both given in magnitude). The index of water is 1.33.
2. (a) (b) A spherical concave mirror has a radius of curvature of 65 cm. An object is placed 40 cm to the left of the mirror. Calculate the image position and its magnification. Sketch a ray diagram clearly illustrating if the image is real or inverted. (6 marks) (i) Define the term critical angle for a light rav nassing from a material with a larger.
VARIANT 23 Suppose an object such as a book page is held 15 cm from a concave lens of focal length - 10.0 cm. Such a lens could be used in eyeglasses to correct pronounced nearsightedness. (a) Using the thin lens equations to calculate the location of the image. (b) What magnification is produced? (c) Use ray tracing to get an approximate location for the image.
4.5-1) The print in many books averages 4.20 mm in height. How high in mm is the image of the print on the retina when the book is held 34.0 cm from the eye? (Assume the lens-to-retina distance is 2.00 cm. Include the sign of the value in your answer.) _______mm
(i) Solve the problem below: 5.47 Compute the image location and magnification of an object 30 cm from the front doublet of the thin-lens combination in Fig. P.5.47. Do the calculation by finding the effect of each lens sepa- rately. Make a sketch of appropriate rays. 15 cm fj = 40 cm = -28 cm (ii) In addition, determine the effective focal length of the two-lens system and locate the principal planes.
4 ) An object of height 9 cm is placed 25 cm in front of a converging lens of focal length 10 cm. Behind the converging lens, and 20 cm from it, there is a diverging lens of the same focl length. (a 15) Find the location of the final image, in centimeters with respect to the diverging lens. (b, 10') What is the magnification of the final image? (c, 5') Is the final image upright or inverted?
Ch. 35
(10) Draw a ray diagram for a 3.0-cm tall object placed 10.0 cm from a converging lens having a focal length of 15.0 cm. 2f 2f

SEE MORE QUESTIONS