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2 years ago

What is the lens equation?

Physics

2 answers:

Sergeeva-Olga [200]2 years ago

8 0

Answer:

1/V + 1/U = 1/F

Explanation:

Lens equation is

1/V + 1/U = 1/F

F = focal length

V = distance of image from the lens

U = distance of object from the lens

Note –

a) Distance of real object & image is negative and positive in sign respectively

b) Height above the optic axis are positive while height below the optic axis will be taken as negative

Strike441 [17]2 years ago

7 0

The lens equation gives d relation between focal length, object distance n image distance.

1/f = 1/v + 1/u

seldon

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Which equation is correct according to Ohm’s law? Which equation is correct according to Ohm’s law? A.) V = IR B.) I = R/V C.) R

vodomira [7]

Answer:

$V = IR$

Explanation:

Required

Which equation represents ohm's law?

Literally, ohm's law implies that current (I) is directly proportional to voltage (V) and inversely proportional to resistance (R).

Mathematically, this can be represented as:

$I\ \alpha\ \frac{V}{R}$

Convert the expression to an equation

$I\ =\ \frac{V}{R}$

Multiply both sides by R to make V the subject

$I\ * R\ =\ \frac{V}{R} * R$

$I\ * R\ =V$

Reorder

$V = I\ * R$

$V = IR$

<em>Option (a) is correct; Others are not</em>

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2 years ago

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A thin double convex glass lens with an index of 1.56 while surrounded by air has a 10 cm focal length. If it is placed under wa

bearhunter [10]

Explanation:

Formula which holds true for a leans with radii $R_{1}$ and $R_{2}$ and index refraction n is given as follows.

$\frac{1}{f} = (n - 1) [\frac{1}{R_{1}} - \frac{1}{R_{2}}]$

Since, the lens is immersed in liquid with index of refraction $n_{1}$ . Therefore, focal length obeys the following.

$\frac{1}{f_{1}} = \frac{n - n_{1}}{n_{1}} [\frac{1}{R_{1}} - \frac{1}{R_{2}}]$

$\frac{1}{f(n - 1)} = [\frac{1}{R_{1}} - \frac{1}{R_{2}}]$

and, $\frac{n_{1}}{f(n - n_{1})} = \frac{1}{R_{1}} - \frac{1}{R_{2}}$

or, $f_{1} = \frac{fn_{1}(n - 1)}{(n - n_{1})}$

$f_{w} = \frac{10 \times 1.33 \times (1.56 - 1)}{(1.56 - 1.33)}$

= 32.4 cm

Using thin lens equation, we will find the focal length as follows.

$\frac{1}{f} = \frac{1}{s_{o}} + \frac{1}{s_{i}}$

Hence, image distance can be calculated as follows.

$\frac{1}{s_{i}} = \frac{1}{f} - \frac{1}{s_{o}} = \frac{s_{o} - f}{fs_{o}}$

$s_{i} = \frac{fs_{o}}{s_{o} - f}$

$s_{i} = \frac{32.4 \times 100}{100 - 32.4}$

= 47.9 cm

Therefore, we can conclude that the focal length of the lens in water is 47.9 cm.

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2 years ago

Please help!!!

Zolol [24]

Answer:

Explanation:

you basically divide 1200 into 25

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3 years ago

a block measures 3.5 cm long 2.8 cm wide and 1.6 cm deep. the density of the block 2.5 g/cm. calculate the volume of the block.

luda_lava [24]

Volume of a block can be found by: length × width × height. So:

3.5cm × 2.8cm × 1.6cm = 15.68cm^3

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2 years ago

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The half-life of caffeine is 5 hours. If you ingested a 30 oz Big Gulp, how many oz of caffeine is left after one half life? * Y

xeze [42]

Answer:

The amount of caffeine left after one half life of 5 hours is 15 oz.

Explanation:

Half life is the time taken for a radioactive substance to degenerate or decay to half of its original size.

The half life of caffeine is 5 hours. So ingesting a 30 oz, this would be reduced to half of its size after the first 5 hours.

So that:

After one half life of 5 hours, the value of caffeine that would be left is;

$\frac{30}{2}$ = 15 oz

The amount of caffeine left after one half life of 5 hours is 15 oz.

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