Match the corresponding entries of Column-I with Column-II.
(Where \(m\) is the magnification produced by the mirror)

Column-I		Column-II
A.	\(m= -2\)	I.	convex mirror
B.	\(m= -\frac{1}{2}\)	II.	concave mirror
C.	\(m= +2\)	III.	real Image
D.	\(m= +\frac{1}{2}\)	IV.	virtual Image

Codes:

	A	B	C	D
1.	I & III	I & IV	I & II	III & IV
2.	I & IV	II & III	II & IV	II & III
3.	III & IV	II & IV	II & III	I & IV
4.	II & III	II & III	II & IV	I & IV

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 \(\frac{I}{2}\)

A light ray is incident at an angle of \(30^{\circ}\) on a transparent surface separating two media. If the angle of refraction is \(60^{\circ}\), then the critical angle is:
1. \(\sin^{- 1} \left(\frac{1}{\sqrt{3}}\right)\)
2. \(\sin^{- 1} \left(\sqrt{3}\right)\)
3. \(\sin^{- 1} \left(\frac{2}{3}\right)\)
4. \(45^{\circ}\)

In the figure shown the angle made by the light ray with the normal in the medium of refractive index \(\sqrt{2}\) is:
                   
1. \(30^{\circ}\)$$

2. \(60^{\circ}\)$$

3. \(90^{\circ}\)$$

4. None of these

A fish is a little away below the surface of a lake. If the critical angle is \(49^{\circ},\) then the fish could see things above the water surface within an angular range of \(\theta^{\circ}\) where: