40% of the X-rays coming from a Coolidge tube can pass through 0.2 mm thick Al plate. The Anode voltage is increased. Now 40% of the X-rays will pass through Al foil of thickness.

zero

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< 0.2 mm

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=

0.2 mm

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> 0.2 mm

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Solution

The correct option is C > 0.2 mm
As the potential difference is increased, frequency and intensity both increase. Hence, to allow same 40% of waves to pass through, thickness of plate reqd will be more than $0.2 m m .$
So, the answer is option (D).

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40% of the X-rays coming from a Coolidge tube can pass through 0.2 mm thick Al plate. The Anode voltage is increased. Now 40% of the X-rays will pass through Al foil of thickness.

The correct option is C > 0.2 mmAs the potential difference is increased, frequency and intensity both increase. Hence, to allow same 40% of waves to pass through, thickness of plate reqd will be more than 0.2mm.So, the answer is option (D).

The correct option is C > 0.2 mm
As the potential difference is increased, frequency and intensity both increase. Hence, to allow same 40% of waves to pass through, thickness of plate reqd will be more than $0.2 m m .$
So, the answer is option (D).