18 An object is placed to the left of a thin diverging lens

18.

An object is placed to the left of a thin diverging lens of focal length 8.0 cm. The image of the object is upright and 0.33 times the actual height. What is the distance of the object to the lens?

7.5 cm

16 cm

12 cm

6.0 cm

19.

A single lens forms a virtual image of an object. Which type of lens could it be?

The lens must be a diverging lens.

The lens could be either a diverging or a converging lens.

The lens must be a converging lens.

20.

A fish appears to be 9.00 m below the surface of a pond when viewed almost directly above by a fisherman. What is the actual depth of the fish? Let the index of refraction n_water =1.33.

6.77 m

0.08 m

11.97 m

0.15 m

Solution

18. focal length=f=-8 cm

let object distance be u=-d
then magnification is given by

image height/object height=image distance/object distance

hence image distance=-0.33*d

now using lens equation,

we get

(1/image distance)-(1/object distance)=1/focal length

hence object distance=16 cm

second option is correct.

19. for creating virtual image, the lens can be either converging or diverging lens.

20.as we know, speed of light in water=speed of light in air/refraction index

so if there had not been any water , the fish would have appeared at 9 m below surface.

time taken for light to travel, in case of no water=9/speed of light in air

now due to water, let the apparent distance be d.
as time taken for the light is same as before,

d/(speed of light in air/refraction index)=9/speed of light in air

d=9/refraction index=6.75 m

so option 1 is correct.