In total internal reflection when the angle of incidence is equal to the critical angle for the pair of media in contact, what will be the angle of refraction?

1.	\(90^{\circ}\)
2.	\(180^{\circ}\)
3.	\(0^{\circ}\)
4.	equal to the angle of incidence

Two similar thin equi-convex lenses, of focal length \(f\) each, are kept coaxially in contact with each other such that the focal length of the combination is \(F_1\)${\mathrm{}}_{}$. When the space between the two lenses is filled with glycerin which has the same refractive index as that of glass \((\mu = 1.5),\) then the equivalent focal length is \(F_2\)${\mathrm{}}_{}$. The ratio \(F_1:F_2\) will be:
1. \(3:4\)
2. \(2:1\)
3. \(1:2\)
4. \(2:3\)

Pick the wrong statement in the context with a rainbow.

A convex lens and a concave lens, each having the same focal length of 25 cm, are put in contact to form a combination of lenses. The power in dioptres of the combination is:

1. 25

2. 50

3. infinite

4. zero

A microscope is focused on a mark on a piece of paper and then a slab of glass of thickness 3 cm and a refractive index 1.5 is placed over the mark. How should the microscope be moved to get the mark in focus again?

1. 1 cm upward

2. 4.5 cm downward

3. 1 cm downward

4. 2 cm upward

The frequency of a light wave in a material is 2×10¹⁴ Hz and the wavelength is 5000 Å. The refractive index of the material will be:

1. 1.40

2. 1.50

3. 3.00

4. 1.33

A boy is trying to start a fire by focusing sunlight on a piece of paper using an equiconvex lens of a focal length of \(10\) cm. The diameter of the sun is \(1.39\times10^9\) m and its mean distance from the earth is \(1.5\times10^{11}\) m. What is the diameter of the sun's image on the paper?
 

Two thin lenses of focal lengths f₁ and f₂ are in contact and coaxial. The power of the combination is:

1. $\sqrt{\frac{{\mathrm{f}}_1}{{\mathrm{f}}_2}}$

2. $\sqrt{\frac{{\mathrm{f}}_2}{{\mathrm{f}}_1}}$

3. $\frac{{\mathrm{f}}_1+{\mathrm{f}}_2}{{\mathrm{f}}_1{\mathrm{f}}_2}$

4. None of the above

A lens having focal length \(f\) and aperture of diameter \(d\) forms an image of intensity \(I\). An aperture of diameter \(\frac{d}{2}\) in central region of lens is covered by a black paper. The focal length of lens and intensity of the image now will be respectively:
1. \(f\) and \(\frac{I}{4}\)
2.  \(\frac{3f}{4}\) and \(\frac{I}{2}\)
3.  \(f\) and \(\frac{3I}{4}\)
4.  \(\frac{f}{2}\) and \(\frac{I}{2}\)