1
JEE Main 2022 (Online) 28th July Evening Shift
MCQ (Single Correct Answer)
+4
-1
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The power of a lens (biconvex) is $$1.25 \mathrm{~m}^{-1}$$ in particular medium. Refractive index of the lens is 1.5 and radii of curvature are $$20 \mathrm{~cm}$$ and $$40 \mathrm{~cm}$$ respectively. The refractive index of surrounding medium:

A
1.0
B
$$\frac{9}{7}$$
C
$$\frac{3}{2}$$
D
$$\frac{4}{3}$$
2
JEE Main 2022 (Online) 28th July Morning Shift
MCQ (Single Correct Answer)
+4
-1
Change Language

As shown in the figure, after passing through the medium 1 . The speed of light $$v_{2}$$ in medium 2 will be :

$$\left(\right.$$ Given $$\mathrm{c}=3 \times 10^{8} \mathrm{~ms}^{-1}$$ )

JEE Main 2022 (Online) 28th July Morning Shift Physics - Geometrical Optics Question 59 English

A
$$1.0 \times 10^{8} \mathrm{~ms}^{-1}$$
B
$$0.5 \times 10^{8} \mathrm{~ms}^{-1}$$
C
$$1.5 \times 10^{8} \mathrm{~ms}^{-1}$$
D
$$3.0 \times 10^{8} \mathrm{~ms}^{-1}$$
3
JEE Main 2022 (Online) 28th July Morning Shift
MCQ (Single Correct Answer)
+4
-1
Change Language

In normal adujstment, for a refracting telescope, the distance between objective and eye piece is $$30 \mathrm{~cm}$$. The focal length of the objective, when the angular magnification of the telescope is 2 , will be :

A
20 cm
B
30 cm
C
10 cm
D
15 cm
4
JEE Main 2022 (Online) 27th July Morning Shift
MCQ (Single Correct Answer)
+4
-1
Change Language

A microscope was initially placed in air (refractive index 1). It is then immersed in oil (refractive index 2). For a light whose wavelength in air is $$\lambda$$, calculate the change of microscope's resolving power due to oil and choose the correct option.

A
Resolving power will be $$\frac{1}{4}$$ in the oil than it was in the air.
B
Resolving power will be twice in the oil than it was in the air.
C
Resolving power will be four times in the oil than it was in the air.
D
Resolving power will be $$\frac{1}{2}$$ in the oil than it was in the air.
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