Question

1a. Calculate the height of the image formed when a 2.1 cm high object is placed 17 cm away from a lens with focal length -10.5 cm.
(Give your answer in cm)

b. When 720 nm wavelength light is diffracted through a double slit the angle of the first dark fringe on each side of the center of the diffraction pattern is 3.5 degrees. Calculate the spacing between the double slits.
(Give your answer in meters.)

c. Two red bottles are placed 8 cm apart. Using 650 nm as the wavelength of the light, calculate the minimum diameter of a camera lens so that the two bottles are resolved when the camera is used from a distance of 170 meters.  
(Give your answer in meters)

Answer 1

(1a) Focal length of the lens = f = - 10.5 cm.

Object distance = u = - 17 cm.

Hence, image distance v, in cm, is given by :

1 / f = 1 / v - 1 / u

or, 1 / v = 1 / f + 1 / u = ( u + f ) / uf

or, v = uf / ( u + f ) = ( - 17 ) ( -10.5 ) / ( - 17 - 10.5 ) ~ - 6.49.

Hence, magnification of the image formed = v / u = - 6.49 / - 17 ~ 0.382

or, image height / object height = 0.382

or, image height = object height x 0.382 = 2.1 cm x 0.382 ~ 0.8 cm.

Hence, height of the image formed is : 0.8 cm.

(b) Condition for formation of a dark fringe in a double - slit setup is : d sin = ( n + 1 / 2 ) ,

where, n = 1 is the order of the dark fringe,

= 3.5^o is the angular position of that fringe,

= 720 nm = 7.2 x 10^-7 m is wavelength of the light used, and,

d = Seperation between the slits.

Hence, d = ( n + 1 / 2 ) / sin = ( 1 + 1 / 2 ) x 7.2 x 10^-7 m / sin 3.5^o

or, d ~ 1.77 x 10^-5 m.

Hence, spacing between the two slits is : 1.77 x 10^-5 m.