[12]13. An object is located at a distance of 56.0 cm to the left of a convex th

Question

[12]13.   An object is located at a distance of 56.0 cm to the left of a convex thin lens that has a radius of curvature of 15.4 cm. (Note that the sign of the radius of curvature is not given.)   Circle the correct word or number in each pair of parentheses to correctly complete the sentence.
Answer: The image is located at (8.93 cm, 16.4 cm, 25.9 cm, 34.5 cm, 48.6 cm, ____ cm,) to the (left, right) of the lens.
The image is (real, virtual).
                        
[18]14. An object stands on the central axis of a thin spherical lens whose focal length is +2.76 cm. The object is 25.2 cm in front of the lens. Circle the correct answer in each pair of parentheses below and also fill in the blanks. Signs must be put in where indicated.
Answer:       The radius of curvature of the lens is (+ ,   -)    ________ cm.   The image
distance is (+ , -) _________ cm.        Magnification (+ ,   -)___________.   The image is (real, virtual).     The orientation
of the image is (inverted, upright) relative to the object.


[4]16.    Circle the TRUE statement(s).

(b) Snell’s law describes how light is refracted when it goes from one medium to another.   
  
[5]17. The electric and magnetic field components of an electromagnetic wave can be described as E(x, t) = Emax sin (kx - t) and
B(x, t) = Bmax sin (kx - t). Fill in the blanks.
(a)    Emax and Bmax, are called the ___________________ of the components.
               (b)    k = 2/.      kis called the _________________ of the wave.    is called the ________________ of the wave.     
(c)    = 2f       is the ____________________frequency.  
               (d)   in (c) above, f is called the __________________

131____________________________
            M2
[5]18. A solid spherical mass M1 = 3.2 x 105 kg and radius 850 km is concentric with a spherical shell
M2 = 6.2 x 105 kg and radius 670 km as shown.   A point mass m = 48 kg is at a distance of 750 km from                     M1        
the center of the solid sphere.     Determine the gravitational force on mass m.                                                                           m Answer: a. 1.35 x 10-15 N.      b. 1.82 x 10-15 N.       c. 3.14 x 10-15 N.       d. 3.88 x 10-15 N.      
e. 5.34 x 10-15 N.       f. ___________                               

For 19 and 20, use the figure at the right which represents two point charges                                                      
Q1 = -7.88 C and Q2 = +4.99 C, which are separated by a distance of 15.8 cm.                 Q1                          
                                                                                                                                                                                                 Q2
[6]19.   Determine the force on Q2 due to Q1.
Answer:   a. 7.64 N, to the right.       b. 9.65 N, to the left.       c. 11.2 N, to the left.       d. 14.2 N, to the left.       e. 17.9 N, to the right.             f. _______________________

Explanation / Answer

Given Data,

radius of curvature , R = 15.4 cm

                          => f = 15.4/2 = 7.7 cm

object distance ,o = 56 cm

let i be the image distance.

Solution :-

We know from lens eqn

1/f = 1/i + 1/o

i = o x f / ( o - f)

i = 56 x 7.7 / (56 - 7.7)

= 8.93 cm

Hence, Image is located at i = 8.93 cm to the right of the lens. The image is real.

[18]14.

Radius of curvature is 2*f = 2*2.76 = + 5.52 cm

value of image distance

1/f = 1/u + 1/v

1/2.76 = 1/25.2 + 1/v

v = + 3.1 cm

Magnification M = - v / u

                        = -3.1 / 25.2

                        = - 0.123

Therefore image type is real and orientation is inverted & lens type is convex

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