NEET 2016 Phase 2 | Geometrical Optics Question 32 | Physics

NEET 2016 Phase 2

MCQ (Single Correct Answer)

-1

A person can see clearly objects only when they lie between 50 cm and 400 cm from his eyes . In order to increase the maximum distance of distinct vision to infinity, the type and power of the correcting lens, the person has to use. will be

convex, + 2.25 diopter

concave, $$-$$0.25 diopter

concave, $$-$$0.2 diopter

convex, + 0.15 diopter

NEET 2016 Phase 1

MCQ (Single Correct Answer)

-1

Match the corresponding entries of column 1 with column 2. [Where m is the magnification produced by the mirror]

Column 1		Column 2
(A)	m = - 2	(p)	Convex mirror
(B)	m = -$${1 \over 2}$$	(q)	Concave mirror
(C)	m = +2	(r)	Real image
(D)	m = +$${1 \over 2}$$	(s)	Virtual image

A $$ \to $$ p and s; B $$ \to $$ q and r; C $$ \to $$ q and s; D $$ \to $$ q and r

A $$ \to $$ r and s; B $$ \to $$ q and s; C $$ \to $$ q and r; D $$ \to $$ p and s

A $$ \to $$ q and r; B $$ \to $$ q and r; C $$ \to $$ q and s; D $$ \to $$ p and s

A $$ \to $$ p and r; B $$ \to $$ p and s; C $$ \to $$ p and q; D $$ \to $$ r and s

NEET 2016 Phase 1

MCQ (Single Correct Answer)

-1

The angle of incidence for a ray of light at a refracting surface of a prism is 45^o. The angle of prism is 60^o. If the ray suffers minimum deviation through the prism, the angle of minimum deviation and refractive index of the material of the prism respectively, are

$${45^o};\sqrt 2 $$

$${30^o};{1 \over {\sqrt 2 }}$$

$${45^o};{1 \over {\sqrt 2 }}$$

$${30^o};\sqrt 2 $$

NEET 2016 Phase 1

MCQ (Single Correct Answer)

-1

A astronomical telescope has objective and eyepiece of focal lengths 40 cm and 4 cm respectively. To view an object 200 cm away from the objective, the lenses must be separated by a distance

50.0 cm

54.0 cm

37.3 cm

46.0 cm

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