1. A glass rod is submerged in water. The end of the rod has the shape of a hemi

Question

1.

A glass rod is submerged in water. The end of the rod has the shape of a hemisphere of radius 2.2 cm, bulging outward. Point P is in the water along the optical axis at a distance of 5.3 cm in front of the glass rod. When a small object is placed at point P, is its image virtual or real, and where does it form? Let the indices of refraction nglass = 1.56 and     nwater = 1.33.

virtual image, 11 cm inside the glass rod

virtual image, 11 cm in front of the glass rod

virtual image, 17 cm in front of the glass rod

real image, 17 cm inside the glass rod

2.

ray 1 only

rays 1 and 2

ray 3 only

ray 2 only

3.

An object is placed very close to the focal point of a concave spherical mirror, but not precisely at the focal point. Where does the image form?

very close to the focal point

very close to the surface of the mirror

very close to the center of curvature

very far away from the mirror

1.

A glass rod is submerged in water. The end of the rod has the shape of a hemisphere of radius 2.2 cm, bulging outward. Point P is in the water along the optical axis at a distance of 5.3 cm in front of the glass rod. When a small object is placed at point P, is its image virtual or real, and where does it form? Let the indices of refraction nglass = 1.56 and     nwater = 1.33.

virtual image, 11 cm inside the glass rod

virtual image, 11 cm in front of the glass rod

virtual image, 17 cm in front of the glass rod

real image, 17 cm inside the glass rod

2.

An object is positioned outside the focal point of a thin convex lens. A student uses the graphical method to predict the location of the image of the arrow. As shown in the diagram, she draws principal rays to the lens only. When the rays are completed, which will pass through F2, the focal point to the right of the lens?

ray 1 only

rays 1 and 2

ray 3 only

ray 2 only

Explanation / Answer

nw = 1.33

nglass = 1.56

from the refraction of glass equation

na/s + nb/s' = (nb-na)/R

s = object distance = 5.3 cm

s' = image distance

(1.33/5.3) + (1.56/s') = (1.56-1.33)/2.2

s' = -11 cm <-----answer

the image is virtual in the water

virtual image , 11cm infront of the glass rod

-------------------------

ray 1 only

-----------------

from the mirror equation

1/s + 1/s' = 1/f

s' = (sf)/(s-f)

as s is near to f

s' = infinity

very far away from the mirror

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