Question

17. Draw a ray trace diagram showing the image location for a 5 cm tall object placed 8 cm away from a converging lens that has a focal length of 10 cm. Is this image real or virtual? Upright or inverted? Smaller or larger than the original object?

18. A slit of width 0.11 mm is cut in a sheet of metal and illuminated with light from a mercury vapor lamp with a wavelength of 577 nm. A screen is placed 4 meters from the slit. Find the width of the central peak in the diffraction pattern in meters on the screen, i.e. find the distance between the first minima on the left and the first minima on the right.

Answer 1

17)

The ray diagram for converging lens is:

From this ray diagram, Final image is Virtual and upright.

18)

Slit widht is d= 0.11 m

Wavelength of light is

Seperation between slit and screen is D= 4 m

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The condition for first minima is,

           dsin=

And,

    sin= tan=

From the above two equations,

         

Rearrange the equation for y,

           

                =(0.11 m)()/(4 m)

               =1.58675x10^-8 m

So, the width of the central peak is,

            Y= 2y

               =2(1.58675x10^-8 m)

              =3.17x10^-8 m

Therefore, the width of the central peak is 3.17x10^-8 m.