When an object is placed at a distance of 15 cm from a lens?

A tank is filled with water to a height of 12.5 cm. The apparent depth of a needle lying at the bottom of the tank is measured by a microscope to be 9.4 cm. What is the refractive index of water? If water is replaced by a liquid of refractive index 1.63 up to the same height, by what distance would the mircoscope have to be moved to focus on the needle again?

Case I: When tank is filled with water. Given, real depth = 12.5 cm;           apparent depth = 9.4 cm Now, using the formula, 

                    μ = real depthapparent depth

Refractive index,  μ = 12.59.4 = 1.33

Case II: When water in the tank is replaced by another liquid.

apparent depth = real depthμ 

i.e., apparent depth = 12.51.63 = 7.67 cm

Distance through which microscope has to be moved downward is = (9.4 – 7.67) cm = 1.73 cm.

Answer

Hint:We need to know about the relation between the formation of an image and the position of the object for both the diverging and converging lenses. The diverging lens makes the light rays diverge and always form a virtual image, while a converging lens forms a real or virtual image based on the location of the object.Complete step by step solution:A concave or converging lens forms a virtual image when the object is placed at a distance less than one focal length. So, in case the object image has to be virtual, the focal length of the converging lens needs to be greater than the distance of the object from the lens, that is 15 cm.While in case of a diverging lens, which diverges all the light rays passing to it, the image produced is always virtual irrespective of the distance of the object.So, all the statements given are correct.Hence, option d is the correct answer.Additional Information:We should know that for a converging lens, the position of the object determines whether the image will be real or virtual. There are certain positions to be considered if we want to know the location of the image formed.For a converging lens, if the object is far from twice the focal length, the image formed will be real and in between the focal point and twice the focal distance. Similarly, by using a ray diagram, we can determine when the image will be real or virtual.Note:In only one case of the converging lenses, the image formed is virtual and in all other cases, the image is real. So, we need to be careful about choosing the correct option and should consider the position of the object given with respect to the lens.

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